QBManager

Ionic Equilibrium

229438 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1

X_{2} Y, 2 \times 10^{- 9} M^{3}

2

XY, 2 \times 10^{- 6} M^{3}

3

{XY}_{2}, 1 \times 10^{- 9} M^{3}

4

{XY}_{2}, 4 \times 10^{- 9} M^{3}

Explanation:

Exp:
$\begin{aligned} d) : [x] = 1 \times 10^{- 3} \\ [y] = 2 \times 10^{- 3} \\ {XY}_{2} ◻ X^{2 +} + 2 Y^{- 3} \\ Y^{-} \\ K_{sp} = [X^{2 +}] {[Y^{-}]}^{2} \\ = [10^{- 3}] {[2 \times 10^{- 3}]}^{2} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
Corresponding stoichiometry $= 4 \times 10^{- 9} M^{3}$

JEE Main 2020

Ionic Equilibrium

229407 Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use?
$\begin{aligned} K_{sp} (AgCl) = 1.8 \times 10^{- 10} \\ K_{sp} (AgBr) = 5.0 \times 10^{- 13} \\ K_{sp} ({Ag}_{2} {CrO}_{4}) = 2.4 \times 10^{- 12} \end{aligned}$

1

AgCl

2

AgBr

3

{Ag}_{2} {CrO}_{4}

4 None of these

Explanation:

Exp:
$\begin{aligned} For binary salts (like AgCl, AgBr) \\ s = \sqrt{K_{sp}} \\ ∴ Solubility of AgCl = \sqrt{1.8 \times 10^{- 10}} \\ = 1.35 \times 10^{- 7} mol / L \\ Solubility of AgBr = \sqrt{5.0 \times 10^{- 13}} \\ = 7.1 \times 10^{- 7} mol / L \\ = \sqrt[3]{\frac{2.4 \times 10^{- 12}}{4}} \\ = \sqrt[3]{600 \times 10^{- 15}} \\ = 8.44 \times 10^{- 5} mol / L \end{aligned}$

BCECE-2006

Ionic Equilibrium

229408 The value of $Λ_{eq}^{\infty}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are respectively, $149.74, 248.1$ and $126.4 {ohm}^{- 1}$ ${cm}^{2} {eq}^{- 1}$ . The value of $Λ_{eq}^{\infty}$ of ${NH}_{4} OH$ is... $Ω {cm}^{2}$ ${eq}^{- 1}$ .

1 371.44

2 271.44

3 71.44

4 Cannot be predicted from given data

Explanation:

Given, $Λ_{eq}^{\infty} ({NH}_{4} Cl) = 149.74$
$\begin{matrix} Λ_{eq}^{\infty} (NaOH) = 248.1 \\ Λ_{eq}^{\infty} (NaCl) = 126.4 \end{matrix}$
Hence,
$\begin{matrix} Λ_{eq}^{\infty} ({NH}_{4} OH) = Λ_{eq}^{\infty} ({NH}_{4} Cl) + Λ_{eq}^{\infty} (NaOH) - Λ_{eq}^{\infty} (NaCL) \\ = (149.74 + 248.1 - 126.4) \\ Λ_{eq}^{\infty} ({NH}_{4} OH) = 271.44 {ohm}^{- 1} {cm}^{2} {eq}^{- 1} \end{matrix}$

BCECE-2008

Ionic Equilibrium

229409 If $x$ mol $L^{- 1}$ is the solubility of $KA l {({SO}_{4})}_{2}$ then $K_{sp}$ is equal to-

1

x^{3}

2

4 x^{4}

3

x^{4}

4

4 x^{3}

Explanation:

$KAl {({SO}_{4})}_{2}$ ?? $K^{+} + {Al}^{3 +} + 2 {SO}_{4}^{2 -}$
$\begin{aligned} K_{sp} & = [K^{+}] [{Al}^{3 +}] {[{SO}_{4^{2}} 2^{-}]}^{2} \\ = x \times x \times (2 x)^{2} \\ = x^{2} \times 4 x^{2} \\ K_{sp} & = 4 x^{4} \end{aligned}$

BCECE-2011

Ionic Equilibrium

229411 Ionic product of water at $310 K$ is $2.7 \times 10^{- 14}$ . What is the $pH$ of neutral water at this temperature?

1 7.00

2 5.98

3 6.78

4 4.58

Explanation:

$K_{w} = [H^{+}] [OH]$
$2.7 \times 10^{- 14} = [H^{+}] [{OH}^{-}]$
As in neutral $[H^{+}] = [{OH}^{-}]$
$\begin{aligned} [H^{+}] = \sqrt{2.7 \times 10^{- 14}} \\ = 1.6432 \times 10^{- 7} \\ pH = - \log [H^{+}] \\ = - \log (1.643 \times 10^{- 7}) \\ pH = 6.78 \end{aligned}$
The $pH$ of Natural water is 6.78

BCECE-2017

Ionic Equilibrium

229438 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1

X_{2} Y, 2 \times 10^{- 9} M^{3}

2

XY, 2 \times 10^{- 6} M^{3}

3

{XY}_{2}, 1 \times 10^{- 9} M^{3}

4

{XY}_{2}, 4 \times 10^{- 9} M^{3}

Explanation:

Exp:
$\begin{aligned} d) : [x] = 1 \times 10^{- 3} \\ [y] = 2 \times 10^{- 3} \\ {XY}_{2} ◻ X^{2 +} + 2 Y^{- 3} \\ Y^{-} \\ K_{sp} = [X^{2 +}] {[Y^{-}]}^{2} \\ = [10^{- 3}] {[2 \times 10^{- 3}]}^{2} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
Corresponding stoichiometry $= 4 \times 10^{- 9} M^{3}$

JEE Main 2020

Ionic Equilibrium

229407 Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use?
$\begin{aligned} K_{sp} (AgCl) = 1.8 \times 10^{- 10} \\ K_{sp} (AgBr) = 5.0 \times 10^{- 13} \\ K_{sp} ({Ag}_{2} {CrO}_{4}) = 2.4 \times 10^{- 12} \end{aligned}$

1

AgCl

2

AgBr

3

{Ag}_{2} {CrO}_{4}

4 None of these

Explanation:

Exp:
$\begin{aligned} For binary salts (like AgCl, AgBr) \\ s = \sqrt{K_{sp}} \\ ∴ Solubility of AgCl = \sqrt{1.8 \times 10^{- 10}} \\ = 1.35 \times 10^{- 7} mol / L \\ Solubility of AgBr = \sqrt{5.0 \times 10^{- 13}} \\ = 7.1 \times 10^{- 7} mol / L \\ = \sqrt[3]{\frac{2.4 \times 10^{- 12}}{4}} \\ = \sqrt[3]{600 \times 10^{- 15}} \\ = 8.44 \times 10^{- 5} mol / L \end{aligned}$

BCECE-2006

Ionic Equilibrium

229408 The value of $Λ_{eq}^{\infty}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are respectively, $149.74, 248.1$ and $126.4 {ohm}^{- 1}$ ${cm}^{2} {eq}^{- 1}$ . The value of $Λ_{eq}^{\infty}$ of ${NH}_{4} OH$ is... $Ω {cm}^{2}$ ${eq}^{- 1}$ .

1 371.44

2 271.44

3 71.44

4 Cannot be predicted from given data

Explanation:

Given, $Λ_{eq}^{\infty} ({NH}_{4} Cl) = 149.74$
$\begin{matrix} Λ_{eq}^{\infty} (NaOH) = 248.1 \\ Λ_{eq}^{\infty} (NaCl) = 126.4 \end{matrix}$
Hence,
$\begin{matrix} Λ_{eq}^{\infty} ({NH}_{4} OH) = Λ_{eq}^{\infty} ({NH}_{4} Cl) + Λ_{eq}^{\infty} (NaOH) - Λ_{eq}^{\infty} (NaCL) \\ = (149.74 + 248.1 - 126.4) \\ Λ_{eq}^{\infty} ({NH}_{4} OH) = 271.44 {ohm}^{- 1} {cm}^{2} {eq}^{- 1} \end{matrix}$

BCECE-2008

Ionic Equilibrium

229409 If $x$ mol $L^{- 1}$ is the solubility of $KA l {({SO}_{4})}_{2}$ then $K_{sp}$ is equal to-

1

x^{3}

2

4 x^{4}

3

x^{4}

4

4 x^{3}

Explanation:

$KAl {({SO}_{4})}_{2}$ ?? $K^{+} + {Al}^{3 +} + 2 {SO}_{4}^{2 -}$
$\begin{aligned} K_{sp} & = [K^{+}] [{Al}^{3 +}] {[{SO}_{4^{2}} 2^{-}]}^{2} \\ = x \times x \times (2 x)^{2} \\ = x^{2} \times 4 x^{2} \\ K_{sp} & = 4 x^{4} \end{aligned}$

BCECE-2011

Ionic Equilibrium

229411 Ionic product of water at $310 K$ is $2.7 \times 10^{- 14}$ . What is the $pH$ of neutral water at this temperature?

1 7.00

2 5.98

3 6.78

4 4.58

Explanation:

$K_{w} = [H^{+}] [OH]$
$2.7 \times 10^{- 14} = [H^{+}] [{OH}^{-}]$
As in neutral $[H^{+}] = [{OH}^{-}]$
$\begin{aligned} [H^{+}] = \sqrt{2.7 \times 10^{- 14}} \\ = 1.6432 \times 10^{- 7} \\ pH = - \log [H^{+}] \\ = - \log (1.643 \times 10^{- 7}) \\ pH = 6.78 \end{aligned}$
The $pH$ of Natural water is 6.78

BCECE-2017

Ionic Equilibrium

229438 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1

X_{2} Y, 2 \times 10^{- 9} M^{3}

2

XY, 2 \times 10^{- 6} M^{3}

3

{XY}_{2}, 1 \times 10^{- 9} M^{3}

4

{XY}_{2}, 4 \times 10^{- 9} M^{3}

Explanation:

Exp:
$\begin{aligned} d) : [x] = 1 \times 10^{- 3} \\ [y] = 2 \times 10^{- 3} \\ {XY}_{2} ◻ X^{2 +} + 2 Y^{- 3} \\ Y^{-} \\ K_{sp} = [X^{2 +}] {[Y^{-}]}^{2} \\ = [10^{- 3}] {[2 \times 10^{- 3}]}^{2} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
Corresponding stoichiometry $= 4 \times 10^{- 9} M^{3}$

JEE Main 2020

Ionic Equilibrium

229407 Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use?
$\begin{aligned} K_{sp} (AgCl) = 1.8 \times 10^{- 10} \\ K_{sp} (AgBr) = 5.0 \times 10^{- 13} \\ K_{sp} ({Ag}_{2} {CrO}_{4}) = 2.4 \times 10^{- 12} \end{aligned}$

1

AgCl

2

AgBr

3

{Ag}_{2} {CrO}_{4}

4 None of these

Explanation:

Exp:
$\begin{aligned} For binary salts (like AgCl, AgBr) \\ s = \sqrt{K_{sp}} \\ ∴ Solubility of AgCl = \sqrt{1.8 \times 10^{- 10}} \\ = 1.35 \times 10^{- 7} mol / L \\ Solubility of AgBr = \sqrt{5.0 \times 10^{- 13}} \\ = 7.1 \times 10^{- 7} mol / L \\ = \sqrt[3]{\frac{2.4 \times 10^{- 12}}{4}} \\ = \sqrt[3]{600 \times 10^{- 15}} \\ = 8.44 \times 10^{- 5} mol / L \end{aligned}$

BCECE-2006

Ionic Equilibrium

229408 The value of $Λ_{eq}^{\infty}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are respectively, $149.74, 248.1$ and $126.4 {ohm}^{- 1}$ ${cm}^{2} {eq}^{- 1}$ . The value of $Λ_{eq}^{\infty}$ of ${NH}_{4} OH$ is... $Ω {cm}^{2}$ ${eq}^{- 1}$ .

1 371.44

2 271.44

3 71.44

4 Cannot be predicted from given data

Explanation:

Given, $Λ_{eq}^{\infty} ({NH}_{4} Cl) = 149.74$
$\begin{matrix} Λ_{eq}^{\infty} (NaOH) = 248.1 \\ Λ_{eq}^{\infty} (NaCl) = 126.4 \end{matrix}$
Hence,
$\begin{matrix} Λ_{eq}^{\infty} ({NH}_{4} OH) = Λ_{eq}^{\infty} ({NH}_{4} Cl) + Λ_{eq}^{\infty} (NaOH) - Λ_{eq}^{\infty} (NaCL) \\ = (149.74 + 248.1 - 126.4) \\ Λ_{eq}^{\infty} ({NH}_{4} OH) = 271.44 {ohm}^{- 1} {cm}^{2} {eq}^{- 1} \end{matrix}$

BCECE-2008

Ionic Equilibrium

229409 If $x$ mol $L^{- 1}$ is the solubility of $KA l {({SO}_{4})}_{2}$ then $K_{sp}$ is equal to-

1

x^{3}

2

4 x^{4}

3

x^{4}

4

4 x^{3}

Explanation:

$KAl {({SO}_{4})}_{2}$ ?? $K^{+} + {Al}^{3 +} + 2 {SO}_{4}^{2 -}$
$\begin{aligned} K_{sp} & = [K^{+}] [{Al}^{3 +}] {[{SO}_{4^{2}} 2^{-}]}^{2} \\ = x \times x \times (2 x)^{2} \\ = x^{2} \times 4 x^{2} \\ K_{sp} & = 4 x^{4} \end{aligned}$

BCECE-2011

Ionic Equilibrium

229411 Ionic product of water at $310 K$ is $2.7 \times 10^{- 14}$ . What is the $pH$ of neutral water at this temperature?

1 7.00

2 5.98

3 6.78

4 4.58

Explanation:

$K_{w} = [H^{+}] [OH]$
$2.7 \times 10^{- 14} = [H^{+}] [{OH}^{-}]$
As in neutral $[H^{+}] = [{OH}^{-}]$
$\begin{aligned} [H^{+}] = \sqrt{2.7 \times 10^{- 14}} \\ = 1.6432 \times 10^{- 7} \\ pH = - \log [H^{+}] \\ = - \log (1.643 \times 10^{- 7}) \\ pH = 6.78 \end{aligned}$
The $pH$ of Natural water is 6.78

BCECE-2017

Ionic Equilibrium

229438 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1

X_{2} Y, 2 \times 10^{- 9} M^{3}

2

XY, 2 \times 10^{- 6} M^{3}

3

{XY}_{2}, 1 \times 10^{- 9} M^{3}

4

{XY}_{2}, 4 \times 10^{- 9} M^{3}

Explanation:

Exp:
$\begin{aligned} d) : [x] = 1 \times 10^{- 3} \\ [y] = 2 \times 10^{- 3} \\ {XY}_{2} ◻ X^{2 +} + 2 Y^{- 3} \\ Y^{-} \\ K_{sp} = [X^{2 +}] {[Y^{-}]}^{2} \\ = [10^{- 3}] {[2 \times 10^{- 3}]}^{2} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
Corresponding stoichiometry $= 4 \times 10^{- 9} M^{3}$

JEE Main 2020

Ionic Equilibrium

229407 Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use?
$\begin{aligned} K_{sp} (AgCl) = 1.8 \times 10^{- 10} \\ K_{sp} (AgBr) = 5.0 \times 10^{- 13} \\ K_{sp} ({Ag}_{2} {CrO}_{4}) = 2.4 \times 10^{- 12} \end{aligned}$

1

AgCl

2

AgBr

3

{Ag}_{2} {CrO}_{4}

4 None of these

Explanation:

Exp:
$\begin{aligned} For binary salts (like AgCl, AgBr) \\ s = \sqrt{K_{sp}} \\ ∴ Solubility of AgCl = \sqrt{1.8 \times 10^{- 10}} \\ = 1.35 \times 10^{- 7} mol / L \\ Solubility of AgBr = \sqrt{5.0 \times 10^{- 13}} \\ = 7.1 \times 10^{- 7} mol / L \\ = \sqrt[3]{\frac{2.4 \times 10^{- 12}}{4}} \\ = \sqrt[3]{600 \times 10^{- 15}} \\ = 8.44 \times 10^{- 5} mol / L \end{aligned}$

BCECE-2006

Ionic Equilibrium

229408 The value of $Λ_{eq}^{\infty}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are respectively, $149.74, 248.1$ and $126.4 {ohm}^{- 1}$ ${cm}^{2} {eq}^{- 1}$ . The value of $Λ_{eq}^{\infty}$ of ${NH}_{4} OH$ is... $Ω {cm}^{2}$ ${eq}^{- 1}$ .

1 371.44

2 271.44

3 71.44

4 Cannot be predicted from given data

Explanation:

Given, $Λ_{eq}^{\infty} ({NH}_{4} Cl) = 149.74$
$\begin{matrix} Λ_{eq}^{\infty} (NaOH) = 248.1 \\ Λ_{eq}^{\infty} (NaCl) = 126.4 \end{matrix}$
Hence,
$\begin{matrix} Λ_{eq}^{\infty} ({NH}_{4} OH) = Λ_{eq}^{\infty} ({NH}_{4} Cl) + Λ_{eq}^{\infty} (NaOH) - Λ_{eq}^{\infty} (NaCL) \\ = (149.74 + 248.1 - 126.4) \\ Λ_{eq}^{\infty} ({NH}_{4} OH) = 271.44 {ohm}^{- 1} {cm}^{2} {eq}^{- 1} \end{matrix}$

BCECE-2008

Ionic Equilibrium

229409 If $x$ mol $L^{- 1}$ is the solubility of $KA l {({SO}_{4})}_{2}$ then $K_{sp}$ is equal to-

1

x^{3}

2

4 x^{4}

3

x^{4}

4

4 x^{3}

Explanation:

$KAl {({SO}_{4})}_{2}$ ?? $K^{+} + {Al}^{3 +} + 2 {SO}_{4}^{2 -}$
$\begin{aligned} K_{sp} & = [K^{+}] [{Al}^{3 +}] {[{SO}_{4^{2}} 2^{-}]}^{2} \\ = x \times x \times (2 x)^{2} \\ = x^{2} \times 4 x^{2} \\ K_{sp} & = 4 x^{4} \end{aligned}$

BCECE-2011

Ionic Equilibrium

229411 Ionic product of water at $310 K$ is $2.7 \times 10^{- 14}$ . What is the $pH$ of neutral water at this temperature?

1 7.00

2 5.98

3 6.78

4 4.58

Explanation:

$K_{w} = [H^{+}] [OH]$
$2.7 \times 10^{- 14} = [H^{+}] [{OH}^{-}]$
As in neutral $[H^{+}] = [{OH}^{-}]$
$\begin{aligned} [H^{+}] = \sqrt{2.7 \times 10^{- 14}} \\ = 1.6432 \times 10^{- 7} \\ pH = - \log [H^{+}] \\ = - \log (1.643 \times 10^{- 7}) \\ pH = 6.78 \end{aligned}$
The $pH$ of Natural water is 6.78

BCECE-2017

Ionic Equilibrium

229438 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

1

X_{2} Y, 2 \times 10^{- 9} M^{3}

2

XY, 2 \times 10^{- 6} M^{3}

3

{XY}_{2}, 1 \times 10^{- 9} M^{3}

4

{XY}_{2}, 4 \times 10^{- 9} M^{3}

Explanation:

Exp:
$\begin{aligned} d) : [x] = 1 \times 10^{- 3} \\ [y] = 2 \times 10^{- 3} \\ {XY}_{2} ◻ X^{2 +} + 2 Y^{- 3} \\ Y^{-} \\ K_{sp} = [X^{2 +}] {[Y^{-}]}^{2} \\ = [10^{- 3}] {[2 \times 10^{- 3}]}^{2} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
Corresponding stoichiometry $= 4 \times 10^{- 9} M^{3}$

JEE Main 2020

Ionic Equilibrium

229407 Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use?
$\begin{aligned} K_{sp} (AgCl) = 1.8 \times 10^{- 10} \\ K_{sp} (AgBr) = 5.0 \times 10^{- 13} \\ K_{sp} ({Ag}_{2} {CrO}_{4}) = 2.4 \times 10^{- 12} \end{aligned}$

1

AgCl

2

AgBr

3

{Ag}_{2} {CrO}_{4}

4 None of these

Explanation:

Exp:
$\begin{aligned} For binary salts (like AgCl, AgBr) \\ s = \sqrt{K_{sp}} \\ ∴ Solubility of AgCl = \sqrt{1.8 \times 10^{- 10}} \\ = 1.35 \times 10^{- 7} mol / L \\ Solubility of AgBr = \sqrt{5.0 \times 10^{- 13}} \\ = 7.1 \times 10^{- 7} mol / L \\ = \sqrt[3]{\frac{2.4 \times 10^{- 12}}{4}} \\ = \sqrt[3]{600 \times 10^{- 15}} \\ = 8.44 \times 10^{- 5} mol / L \end{aligned}$

BCECE-2006

Ionic Equilibrium

229408 The value of $Λ_{eq}^{\infty}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are respectively, $149.74, 248.1$ and $126.4 {ohm}^{- 1}$ ${cm}^{2} {eq}^{- 1}$ . The value of $Λ_{eq}^{\infty}$ of ${NH}_{4} OH$ is... $Ω {cm}^{2}$ ${eq}^{- 1}$ .

1 371.44

2 271.44

3 71.44

4 Cannot be predicted from given data

Explanation:

Given, $Λ_{eq}^{\infty} ({NH}_{4} Cl) = 149.74$
$\begin{matrix} Λ_{eq}^{\infty} (NaOH) = 248.1 \\ Λ_{eq}^{\infty} (NaCl) = 126.4 \end{matrix}$
Hence,
$\begin{matrix} Λ_{eq}^{\infty} ({NH}_{4} OH) = Λ_{eq}^{\infty} ({NH}_{4} Cl) + Λ_{eq}^{\infty} (NaOH) - Λ_{eq}^{\infty} (NaCL) \\ = (149.74 + 248.1 - 126.4) \\ Λ_{eq}^{\infty} ({NH}_{4} OH) = 271.44 {ohm}^{- 1} {cm}^{2} {eq}^{- 1} \end{matrix}$

BCECE-2008

Ionic Equilibrium

229409 If $x$ mol $L^{- 1}$ is the solubility of $KA l {({SO}_{4})}_{2}$ then $K_{sp}$ is equal to-

1

x^{3}

2

4 x^{4}

3

x^{4}

4

4 x^{3}

Explanation:

$KAl {({SO}_{4})}_{2}$ ?? $K^{+} + {Al}^{3 +} + 2 {SO}_{4}^{2 -}$
$\begin{aligned} K_{sp} & = [K^{+}] [{Al}^{3 +}] {[{SO}_{4^{2}} 2^{-}]}^{2} \\ = x \times x \times (2 x)^{2} \\ = x^{2} \times 4 x^{2} \\ K_{sp} & = 4 x^{4} \end{aligned}$

BCECE-2011

Ionic Equilibrium

229411 Ionic product of water at $310 K$ is $2.7 \times 10^{- 14}$ . What is the $pH$ of neutral water at this temperature?

1 7.00

2 5.98

3 6.78

4 4.58

Explanation:

$K_{w} = [H^{+}] [OH]$
$2.7 \times 10^{- 14} = [H^{+}] [{OH}^{-}]$
As in neutral $[H^{+}] = [{OH}^{-}]$
$\begin{aligned} [H^{+}] = \sqrt{2.7 \times 10^{- 14}} \\ = 1.6432 \times 10^{- 7} \\ pH = - \log [H^{+}] \\ = - \log (1.643 \times 10^{- 7}) \\ pH = 6.78 \end{aligned}$
The $pH$ of Natural water is 6.78

BCECE-2017

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