pH Calculations for Different Solutions

Equal volumes of $0.5\mathrm{M} \mathrm{NaOH}$ and $0.3\mathrm{M} \mathrm{KOH}$ are mixed in an experiment. Find the $\mathrm{pH}$ of the resulting solution.

For a weak acid $\mathrm{HA}$, the value of ${\mathrm{K}}_{\mathrm{a}}={10}^{-4},$ then calculate the $\mathrm{pH}$ of $0.01 \mathrm{M}$ solution of this acid.

When $\text{100}\hspace{0.33em}\mathrm{m}\mathrm{l}$ of $\text{0.2}\hspace{0.33em}\mathrm{M}\hspace{0.33em}{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ is added to $\text{100}\hspace{0.33em}\mathrm{m}\mathrm{l}$ of $\text{0.2}\hspace{0.33em}\mathrm{M}\hspace{0.33em}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}.$ What is the nature of the resulting solution obtained?

The $\mathrm{p}\mathrm{H}$ at the equivalence point for the titration of $0.10\mathrm{ }\mathrm{M}$ $\mathrm{K}{\mathrm{H}}_2\mathrm{B}{\mathrm{O}}_3$ with $0.1\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}$ is (Report the answer in the nearest integer value)
$\left({\mathrm{K}}_{\mathrm{a}}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}{\mathrm{H}}_{\mathrm{3}}{\mathrm{BO}}_{\mathrm{3}}=12.8{\times 10}^{-\mathrm{10}}\right)$

If the concentration of $\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]$ in a solution having $0.02\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{N}{\mathrm{H}}_3$ and $0.005\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{C}\mathrm{a}{\left(\mathrm{O}\mathrm{H}\right)}_2$ is $\mathrm{a}\times {10}^{-6}\mathrm{ }\mathrm{M}$ , determine $\mathrm{a}$ . $[{\mathrm{k}}_{\mathrm{b}}\left(\mathrm{N}{\mathrm{H}}_3\right)=1.8\times {10}^{-5}]$

$10 ml$ of $10 M$ $H_{2}S{O}_4$ is mixed to $100 ml$ $1M$ $NaOH$ solution. The resultant solution will be

$\mathrm{p}{{\mathrm{K}}_{\mathrm{a}}}_1\mathrm{ },\mathrm{ }\mathrm{p}{{\mathrm{K}}_{\mathrm{a}}}_2$ and $\mathrm{p}{{\mathrm{K}}_{\mathrm{a}}}_3$ of ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ are respectively x, y and z. pH of 0.01 M $\mathrm{N}{\mathrm{a}}_2\mathrm{H}\mathrm{P}{\mathrm{O}}_4$ solution is:

A $90\mathrm{ }\mathrm{m}\mathrm{l}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}$ solution of $0.1\mathrm{M}$ is added to $10\mathrm{ }\mathrm{m}\mathrm{L}$ $0.89\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ solution. To this mixture, $40\mathrm{ }\mathrm{m}\mathrm{L}$ of ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ solution is added to get a $\mathrm{p}\mathrm{H}$ of 2. The molarity of ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ added is -

The pH of a solution obtained by mixing 50 mL of 1 N $\mathrm{H}\mathrm{C}\mathrm{l}$ and 30 mL of 1 N $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ is [$\log 2.5=0.3979$]

A solution contains $10 \mathrm{mL} 0.1 \mathrm{N}$ $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ and $10 \mathrm{mL} 0.05 \mathrm{N}$ ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$. What is the ${\mathrm{p}}^{\mathrm{H}}$ of the corresponding solution?

The pH of the solution

$5\mathrm{ml} \frac{\mathrm{M}}{5}\mathrm{ }\mathrm{HCl}+10\mathrm{ }\mathrm{mL}\mathrm{ }\mathrm{of}\mathrm{ }\frac{\mathrm{M}}{10}\mathrm{ }\mathrm{NaOH}$ is

Mark out the incorrect statements :

$$\begin{gathered} \begin{array}{ccccc}{\mathrm{H}}_3{\mathrm{PO}}_4& \rightleftharpoons & {\mathrm{H}}^{\oplus }+{\mathrm{H}}_2{\mathrm{PO}}_4^{\ominus }& \mathrm{;}& {\mathrm{K}}_{{\mathrm{a}}_1}:\end{array}{10}^{-3} \\ \begin{array}{ccccc}{\mathrm{H}}_2{\mathrm{PO}}_4^{\ominus }& \rightleftharpoons & {\mathrm{H}}^{\oplus }+{\mathrm{HPO}}_4^{2-}& ;& {\mathrm{K}}_{{\mathrm{a}}_2}:\end{array} {10}^{-7} \\ \begin{array}{ccccc}{\mathrm{HPO}}_4^{2-}& \rightleftharpoons & {\mathrm{H}}^{\oplus }+{\mathrm{PO}}_4^{3-}& ;& {\mathrm{K}}_{{\mathrm{a}}_3}:\end{array} {10}^{-13} \end{gathered}$$

A solution was prepared by mixing $50 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $50 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$. The $\mathrm{pH}$ of the solution is

A solution was prepared by mixing $50 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $50 \mathrm{mL} 0.10 \mathrm{M} \mathrm{NaOH}$. What is the $\mathrm{pH}$ of the solution?

What will be the $\mathrm{pH}$ of the solution on mixing two equal volume of solutions having $\mathrm{pH}=3$ and $\mathrm{pH}=4$? $\left[\log 5.5=0.7404\right]$

Bromine in excess is dropped to a $0.01\text{M}$ ${\mathrm{SO}}_2$. All of ${\mathrm{SO}}_2$ is oxidized to ${\mathrm{H}}_2{\mathrm{SO}}_4$ and the excess ${\mathrm{Br}}_2$ is removed by flushing with gaseous ${\mathrm{N}}_2$. Determine the $\text{pH}$ of the resulting solution assuming ${\mathrm{K}}_{\mathrm{a}1}$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$ very large & ${\text{K}}_{\text{a2}}={10}^{-2}.$ Take the value of $\log (3.24) = 0.51$. Give the answer to the nearest integer value after multiplying with 100.

Which kind of attractive forces are likely to be holding particles together in a substance that melts at $681 °\mathrm{C}$ and that conducts electricity when molten but not when solid?

A solution is prepared by mixing 100 mL 0.50 M hydrazoic acid (HN₃), whose K_a = 3.6 x 10^{- 4} , with 400 mL of 0.10 M cyanic acid (HOCN), whose K_a = 8 x 10^{- 4} . Which of the following is (are) true regarding.

Which of the following solutions will have pH close to 1.0 ?

$10 \mathrm{mL}$ of $0.1 \mathrm{M}$ tribasic acid ${\mathrm{H}}_3\mathrm{A}$ is titrated with  $0.1 \mathrm{M} \mathrm{NaOH}$ solution. What is the ratio of $\frac{\left[{\text{H}}_3\text{A}\right]}{\left[{\text{A}}^{3-}\right]}$ at $2^{\mathrm{nd}}$ equivalence point? $\left[ {\mathrm{K}}_{{\mathrm{a}}_1}={10}^{-3}, {\mathrm{K}}_{{\mathrm{a}}_2}={10}^{-8}, {\mathrm{K}}_{{\mathrm{a}}_3}={10}^{-12}\right]$