229799
Calculate the molar ratio of a weak acid HA $\left(K_a=10^{-6}\right)$ and its salt with strong base, so that the $\mathrm{pH}$ of buffer solution is 6 .
1 10
2 1
3 6
4 0.1
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$ $\begin{gathered} =\frac{\mathrm{pH}}{\mathrm{pK}_{\mathrm{a}}}=\frac{6}{6}=1 \\ {\left[\mathrm{~K}_{\mathrm{a}}=10^{-6} \therefore \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\ \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\ \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \\ \frac{[\text { Acid }]}{[\text { Salt }]}=0.1 \end{gathered}$ Hence, the correct option is (d).
Shift-II
Ionic Equilibrium
229800
$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{ml}$ of potassium acetate. $\mathrm{Ka}$ of acetic acid = $1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C}$ calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is 4.8
1 $0.1 \mathrm{M}$
2 $0.04 \mathrm{M}$
3 $0.4 \mathrm{M}$
4 $0.02 \mathrm{M}$
Explanation:
Let, the concentration of potassium acetate is $\mathrm{x}$. According to Henderson's equation, $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{\mathrm{x} \times 50}{20 \times 0.1} \\ & 4.8=5 \log _{10} 10-\log 1.8+\log 25 \mathrm{x} \\ & 4.8=5-0.2552+\log 25 \mathrm{x} \\ & 4.8=4.74+\log 25 \mathrm{x} \\ & \text { or } \log 25 \mathrm{x}=0.06 \\ & 25 \mathrm{x}=1.48 \\ & \therefore \mathrm{x}=0.045 \mathrm{M} \end{aligned}$
AP EAMCET (Engg.)-2009
Ionic Equilibrium
229801
$\mathrm{pH}$ of a buffer solution decreases by 0.02 units when $0.12 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} \mathrm{C}$. The buffer capacity of the solution is
1 0.1
2 10
3 1
4 0.4
Explanation:
Buffer capacity, $\beta=\frac{\mathrm{dC}_{\mathrm{HH}-}}{\mathrm{d}_{\mathrm{pH}}}$ where, $\mathrm{dC}_{\mathrm{HA}}=$ no. of moles of acid added per litre $\mathrm{d}_{\mathrm{pH}}=$ change in $\mathrm{pH}$ $\begin{aligned} & \mathrm{dC}_{\mathrm{HA}}= \frac{\text { Moles of acetic acid }}{\text { Volume }} \\ &=\frac{0.12 / 60}{250 / 1000}=\frac{0.12 \times 1000}{60 \times 250}=\frac{0.12 \times 4}{60}=\frac{0.12}{15} \\ & \beta=\frac{0.12}{0.02 \times 15}=\frac{6}{15}=0.4 \end{aligned}$
229799
Calculate the molar ratio of a weak acid HA $\left(K_a=10^{-6}\right)$ and its salt with strong base, so that the $\mathrm{pH}$ of buffer solution is 6 .
1 10
2 1
3 6
4 0.1
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$ $\begin{gathered} =\frac{\mathrm{pH}}{\mathrm{pK}_{\mathrm{a}}}=\frac{6}{6}=1 \\ {\left[\mathrm{~K}_{\mathrm{a}}=10^{-6} \therefore \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\ \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\ \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \\ \frac{[\text { Acid }]}{[\text { Salt }]}=0.1 \end{gathered}$ Hence, the correct option is (d).
Shift-II
Ionic Equilibrium
229800
$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{ml}$ of potassium acetate. $\mathrm{Ka}$ of acetic acid = $1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C}$ calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is 4.8
1 $0.1 \mathrm{M}$
2 $0.04 \mathrm{M}$
3 $0.4 \mathrm{M}$
4 $0.02 \mathrm{M}$
Explanation:
Let, the concentration of potassium acetate is $\mathrm{x}$. According to Henderson's equation, $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{\mathrm{x} \times 50}{20 \times 0.1} \\ & 4.8=5 \log _{10} 10-\log 1.8+\log 25 \mathrm{x} \\ & 4.8=5-0.2552+\log 25 \mathrm{x} \\ & 4.8=4.74+\log 25 \mathrm{x} \\ & \text { or } \log 25 \mathrm{x}=0.06 \\ & 25 \mathrm{x}=1.48 \\ & \therefore \mathrm{x}=0.045 \mathrm{M} \end{aligned}$
AP EAMCET (Engg.)-2009
Ionic Equilibrium
229801
$\mathrm{pH}$ of a buffer solution decreases by 0.02 units when $0.12 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} \mathrm{C}$. The buffer capacity of the solution is
1 0.1
2 10
3 1
4 0.4
Explanation:
Buffer capacity, $\beta=\frac{\mathrm{dC}_{\mathrm{HH}-}}{\mathrm{d}_{\mathrm{pH}}}$ where, $\mathrm{dC}_{\mathrm{HA}}=$ no. of moles of acid added per litre $\mathrm{d}_{\mathrm{pH}}=$ change in $\mathrm{pH}$ $\begin{aligned} & \mathrm{dC}_{\mathrm{HA}}= \frac{\text { Moles of acetic acid }}{\text { Volume }} \\ &=\frac{0.12 / 60}{250 / 1000}=\frac{0.12 \times 1000}{60 \times 250}=\frac{0.12 \times 4}{60}=\frac{0.12}{15} \\ & \beta=\frac{0.12}{0.02 \times 15}=\frac{6}{15}=0.4 \end{aligned}$
229799
Calculate the molar ratio of a weak acid HA $\left(K_a=10^{-6}\right)$ and its salt with strong base, so that the $\mathrm{pH}$ of buffer solution is 6 .
1 10
2 1
3 6
4 0.1
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$ $\begin{gathered} =\frac{\mathrm{pH}}{\mathrm{pK}_{\mathrm{a}}}=\frac{6}{6}=1 \\ {\left[\mathrm{~K}_{\mathrm{a}}=10^{-6} \therefore \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\ \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\ \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \\ \frac{[\text { Acid }]}{[\text { Salt }]}=0.1 \end{gathered}$ Hence, the correct option is (d).
Shift-II
Ionic Equilibrium
229800
$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{ml}$ of potassium acetate. $\mathrm{Ka}$ of acetic acid = $1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C}$ calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is 4.8
1 $0.1 \mathrm{M}$
2 $0.04 \mathrm{M}$
3 $0.4 \mathrm{M}$
4 $0.02 \mathrm{M}$
Explanation:
Let, the concentration of potassium acetate is $\mathrm{x}$. According to Henderson's equation, $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{\mathrm{x} \times 50}{20 \times 0.1} \\ & 4.8=5 \log _{10} 10-\log 1.8+\log 25 \mathrm{x} \\ & 4.8=5-0.2552+\log 25 \mathrm{x} \\ & 4.8=4.74+\log 25 \mathrm{x} \\ & \text { or } \log 25 \mathrm{x}=0.06 \\ & 25 \mathrm{x}=1.48 \\ & \therefore \mathrm{x}=0.045 \mathrm{M} \end{aligned}$
AP EAMCET (Engg.)-2009
Ionic Equilibrium
229801
$\mathrm{pH}$ of a buffer solution decreases by 0.02 units when $0.12 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} \mathrm{C}$. The buffer capacity of the solution is
1 0.1
2 10
3 1
4 0.4
Explanation:
Buffer capacity, $\beta=\frac{\mathrm{dC}_{\mathrm{HH}-}}{\mathrm{d}_{\mathrm{pH}}}$ where, $\mathrm{dC}_{\mathrm{HA}}=$ no. of moles of acid added per litre $\mathrm{d}_{\mathrm{pH}}=$ change in $\mathrm{pH}$ $\begin{aligned} & \mathrm{dC}_{\mathrm{HA}}= \frac{\text { Moles of acetic acid }}{\text { Volume }} \\ &=\frac{0.12 / 60}{250 / 1000}=\frac{0.12 \times 1000}{60 \times 250}=\frac{0.12 \times 4}{60}=\frac{0.12}{15} \\ & \beta=\frac{0.12}{0.02 \times 15}=\frac{6}{15}=0.4 \end{aligned}$
229799
Calculate the molar ratio of a weak acid HA $\left(K_a=10^{-6}\right)$ and its salt with strong base, so that the $\mathrm{pH}$ of buffer solution is 6 .
1 10
2 1
3 6
4 0.1
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$ $\begin{gathered} =\frac{\mathrm{pH}}{\mathrm{pK}_{\mathrm{a}}}=\frac{6}{6}=1 \\ {\left[\mathrm{~K}_{\mathrm{a}}=10^{-6} \therefore \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\ \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\ \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \\ \frac{[\text { Acid }]}{[\text { Salt }]}=0.1 \end{gathered}$ Hence, the correct option is (d).
Shift-II
Ionic Equilibrium
229800
$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{ml}$ of potassium acetate. $\mathrm{Ka}$ of acetic acid = $1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C}$ calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is 4.8
1 $0.1 \mathrm{M}$
2 $0.04 \mathrm{M}$
3 $0.4 \mathrm{M}$
4 $0.02 \mathrm{M}$
Explanation:
Let, the concentration of potassium acetate is $\mathrm{x}$. According to Henderson's equation, $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{\mathrm{x} \times 50}{20 \times 0.1} \\ & 4.8=5 \log _{10} 10-\log 1.8+\log 25 \mathrm{x} \\ & 4.8=5-0.2552+\log 25 \mathrm{x} \\ & 4.8=4.74+\log 25 \mathrm{x} \\ & \text { or } \log 25 \mathrm{x}=0.06 \\ & 25 \mathrm{x}=1.48 \\ & \therefore \mathrm{x}=0.045 \mathrm{M} \end{aligned}$
AP EAMCET (Engg.)-2009
Ionic Equilibrium
229801
$\mathrm{pH}$ of a buffer solution decreases by 0.02 units when $0.12 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} \mathrm{C}$. The buffer capacity of the solution is
1 0.1
2 10
3 1
4 0.4
Explanation:
Buffer capacity, $\beta=\frac{\mathrm{dC}_{\mathrm{HH}-}}{\mathrm{d}_{\mathrm{pH}}}$ where, $\mathrm{dC}_{\mathrm{HA}}=$ no. of moles of acid added per litre $\mathrm{d}_{\mathrm{pH}}=$ change in $\mathrm{pH}$ $\begin{aligned} & \mathrm{dC}_{\mathrm{HA}}= \frac{\text { Moles of acetic acid }}{\text { Volume }} \\ &=\frac{0.12 / 60}{250 / 1000}=\frac{0.12 \times 1000}{60 \times 250}=\frac{0.12 \times 4}{60}=\frac{0.12}{15} \\ & \beta=\frac{0.12}{0.02 \times 15}=\frac{6}{15}=0.4 \end{aligned}$
