229773
If in $100 \mathrm{ml}$ of an aqueous HCl of $\mathrm{pH} 1.00,900$ $\mathrm{ml}$ of more distilled water are added, the $\mathrm{pH}$ of the resultant solution will be
1 1.0
2 2.0
3 4.0
4 7.0
Explanation:
Given an aqueous $\mathrm{HCl}$ of $\mathrm{pH}=1$ we know that- $\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=1 \\ & \mathrm{H}^{+}=10^{-1} \end{aligned}$ When $900 \mathrm{ml}$ water is added to $100 \mathrm{ml}$ solution, total volume becomes $1000 \mathrm{ml}$. Thus the solution is diluted to 10 times. Hence the concentration will be one tenth of original concentration. $\left[\mathrm{H}^{+}\right]=10^{-2}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-2}=2.0$
AP-EAMCET-1992
Ionic Equilibrium
229774
The ionic product of water is $10^{-14}$. What is the hydrogen ion concentration of a $0.1 \mathrm{M} \mathrm{NaOH}$ solution?
1 $10^{-13} \mathrm{M}$
2 $10^{-14}$
3 $14 \mathrm{M}$
4 $13 \mathrm{M}$
Explanation:
As we know that, $\mathrm{K}_\pi=\left[\mathrm{H}^{+}\right][\mathrm{OH}]$ Given :- $\begin{aligned} & \text { Given :- } \\ & \mathrm{K}_{\mathrm{W}}=10^{-14} \\ & {[\mathrm{OH}]=0.1=10^{-1}} \\ & \therefore\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13} \mathrm{M} \end{aligned}$
AP-EAMCET-1994
Ionic Equilibrium
229775
The $\mathrm{pH}$ of a $0.05 \mathrm{M}$ solution of $\mathrm{H}_2 \mathrm{SO}_4$ is
229777
Urine normally has a $\mathrm{pH}$ of 6 . If a patient eliminates $1.3 \mathrm{~L}$ of urine per day, how many moles of $\mathrm{H}^{+}$ions does he urinate?
1 $1.3 \times 10^{-3}$
2 $1.3 \times 10^{-6}$
3 $1.3 \times 10^{-7}$
4 $1.3 \times 10^6$
Explanation:
Given that, $\mathrm{pH}$ of urine is $=6$ $\therefore \quad\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$ $\left[\mathrm{H}^{+}\right]=\frac{\text { no. of moles }}{\text { volume }}$ no. of moles $=10^{-6} \times 1.3$ A patient eliminates $1.3 \mathrm{~L}$ of urine per day then $1.3 \times$ $10^{-6} \mathrm{~mol} / \mathrm{lit}$. of $\mathrm{H}^{+}$ions present.
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Ionic Equilibrium
229773
If in $100 \mathrm{ml}$ of an aqueous HCl of $\mathrm{pH} 1.00,900$ $\mathrm{ml}$ of more distilled water are added, the $\mathrm{pH}$ of the resultant solution will be
1 1.0
2 2.0
3 4.0
4 7.0
Explanation:
Given an aqueous $\mathrm{HCl}$ of $\mathrm{pH}=1$ we know that- $\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=1 \\ & \mathrm{H}^{+}=10^{-1} \end{aligned}$ When $900 \mathrm{ml}$ water is added to $100 \mathrm{ml}$ solution, total volume becomes $1000 \mathrm{ml}$. Thus the solution is diluted to 10 times. Hence the concentration will be one tenth of original concentration. $\left[\mathrm{H}^{+}\right]=10^{-2}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-2}=2.0$
AP-EAMCET-1992
Ionic Equilibrium
229774
The ionic product of water is $10^{-14}$. What is the hydrogen ion concentration of a $0.1 \mathrm{M} \mathrm{NaOH}$ solution?
1 $10^{-13} \mathrm{M}$
2 $10^{-14}$
3 $14 \mathrm{M}$
4 $13 \mathrm{M}$
Explanation:
As we know that, $\mathrm{K}_\pi=\left[\mathrm{H}^{+}\right][\mathrm{OH}]$ Given :- $\begin{aligned} & \text { Given :- } \\ & \mathrm{K}_{\mathrm{W}}=10^{-14} \\ & {[\mathrm{OH}]=0.1=10^{-1}} \\ & \therefore\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13} \mathrm{M} \end{aligned}$
AP-EAMCET-1994
Ionic Equilibrium
229775
The $\mathrm{pH}$ of a $0.05 \mathrm{M}$ solution of $\mathrm{H}_2 \mathrm{SO}_4$ is
229777
Urine normally has a $\mathrm{pH}$ of 6 . If a patient eliminates $1.3 \mathrm{~L}$ of urine per day, how many moles of $\mathrm{H}^{+}$ions does he urinate?
1 $1.3 \times 10^{-3}$
2 $1.3 \times 10^{-6}$
3 $1.3 \times 10^{-7}$
4 $1.3 \times 10^6$
Explanation:
Given that, $\mathrm{pH}$ of urine is $=6$ $\therefore \quad\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$ $\left[\mathrm{H}^{+}\right]=\frac{\text { no. of moles }}{\text { volume }}$ no. of moles $=10^{-6} \times 1.3$ A patient eliminates $1.3 \mathrm{~L}$ of urine per day then $1.3 \times$ $10^{-6} \mathrm{~mol} / \mathrm{lit}$. of $\mathrm{H}^{+}$ions present.
229773
If in $100 \mathrm{ml}$ of an aqueous HCl of $\mathrm{pH} 1.00,900$ $\mathrm{ml}$ of more distilled water are added, the $\mathrm{pH}$ of the resultant solution will be
1 1.0
2 2.0
3 4.0
4 7.0
Explanation:
Given an aqueous $\mathrm{HCl}$ of $\mathrm{pH}=1$ we know that- $\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=1 \\ & \mathrm{H}^{+}=10^{-1} \end{aligned}$ When $900 \mathrm{ml}$ water is added to $100 \mathrm{ml}$ solution, total volume becomes $1000 \mathrm{ml}$. Thus the solution is diluted to 10 times. Hence the concentration will be one tenth of original concentration. $\left[\mathrm{H}^{+}\right]=10^{-2}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-2}=2.0$
AP-EAMCET-1992
Ionic Equilibrium
229774
The ionic product of water is $10^{-14}$. What is the hydrogen ion concentration of a $0.1 \mathrm{M} \mathrm{NaOH}$ solution?
1 $10^{-13} \mathrm{M}$
2 $10^{-14}$
3 $14 \mathrm{M}$
4 $13 \mathrm{M}$
Explanation:
As we know that, $\mathrm{K}_\pi=\left[\mathrm{H}^{+}\right][\mathrm{OH}]$ Given :- $\begin{aligned} & \text { Given :- } \\ & \mathrm{K}_{\mathrm{W}}=10^{-14} \\ & {[\mathrm{OH}]=0.1=10^{-1}} \\ & \therefore\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13} \mathrm{M} \end{aligned}$
AP-EAMCET-1994
Ionic Equilibrium
229775
The $\mathrm{pH}$ of a $0.05 \mathrm{M}$ solution of $\mathrm{H}_2 \mathrm{SO}_4$ is
229777
Urine normally has a $\mathrm{pH}$ of 6 . If a patient eliminates $1.3 \mathrm{~L}$ of urine per day, how many moles of $\mathrm{H}^{+}$ions does he urinate?
1 $1.3 \times 10^{-3}$
2 $1.3 \times 10^{-6}$
3 $1.3 \times 10^{-7}$
4 $1.3 \times 10^6$
Explanation:
Given that, $\mathrm{pH}$ of urine is $=6$ $\therefore \quad\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$ $\left[\mathrm{H}^{+}\right]=\frac{\text { no. of moles }}{\text { volume }}$ no. of moles $=10^{-6} \times 1.3$ A patient eliminates $1.3 \mathrm{~L}$ of urine per day then $1.3 \times$ $10^{-6} \mathrm{~mol} / \mathrm{lit}$. of $\mathrm{H}^{+}$ions present.
229773
If in $100 \mathrm{ml}$ of an aqueous HCl of $\mathrm{pH} 1.00,900$ $\mathrm{ml}$ of more distilled water are added, the $\mathrm{pH}$ of the resultant solution will be
1 1.0
2 2.0
3 4.0
4 7.0
Explanation:
Given an aqueous $\mathrm{HCl}$ of $\mathrm{pH}=1$ we know that- $\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=1 \\ & \mathrm{H}^{+}=10^{-1} \end{aligned}$ When $900 \mathrm{ml}$ water is added to $100 \mathrm{ml}$ solution, total volume becomes $1000 \mathrm{ml}$. Thus the solution is diluted to 10 times. Hence the concentration will be one tenth of original concentration. $\left[\mathrm{H}^{+}\right]=10^{-2}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-2}=2.0$
AP-EAMCET-1992
Ionic Equilibrium
229774
The ionic product of water is $10^{-14}$. What is the hydrogen ion concentration of a $0.1 \mathrm{M} \mathrm{NaOH}$ solution?
1 $10^{-13} \mathrm{M}$
2 $10^{-14}$
3 $14 \mathrm{M}$
4 $13 \mathrm{M}$
Explanation:
As we know that, $\mathrm{K}_\pi=\left[\mathrm{H}^{+}\right][\mathrm{OH}]$ Given :- $\begin{aligned} & \text { Given :- } \\ & \mathrm{K}_{\mathrm{W}}=10^{-14} \\ & {[\mathrm{OH}]=0.1=10^{-1}} \\ & \therefore\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13} \mathrm{M} \end{aligned}$
AP-EAMCET-1994
Ionic Equilibrium
229775
The $\mathrm{pH}$ of a $0.05 \mathrm{M}$ solution of $\mathrm{H}_2 \mathrm{SO}_4$ is
229777
Urine normally has a $\mathrm{pH}$ of 6 . If a patient eliminates $1.3 \mathrm{~L}$ of urine per day, how many moles of $\mathrm{H}^{+}$ions does he urinate?
1 $1.3 \times 10^{-3}$
2 $1.3 \times 10^{-6}$
3 $1.3 \times 10^{-7}$
4 $1.3 \times 10^6$
Explanation:
Given that, $\mathrm{pH}$ of urine is $=6$ $\therefore \quad\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} / \mathrm{L}$ $\left[\mathrm{H}^{+}\right]=\frac{\text { no. of moles }}{\text { volume }}$ no. of moles $=10^{-6} \times 1.3$ A patient eliminates $1.3 \mathrm{~L}$ of urine per day then $1.3 \times$ $10^{-6} \mathrm{~mol} / \mathrm{lit}$. of $\mathrm{H}^{+}$ions present.