Chemistry, 31.07.2019 22:30 marvin07

One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the pka of hno2 is 3.4.

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Answer from: brittneylouise2852

The new pH of given solutionis $\boxed{3.53}$ .

Further Explanation:

The aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is termed as buffer solution. These solutions offer strong resistance to any change in their pH on addition of small quantity of strong acid or base.

Henderson-Hasselbalch equation:

This equation helps in determining the pH of buffer solution. Its mathematical form is given as follows:

${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}$ …… (1)

Here,

$\left[ {{{\text{A}}^ - }} \right]$ is concentration of conjugate base.

[HA] is concentration of acid.

Given mixture is a buffer solution of ${\text{HN}}{{\text{O}}_{\text{2}}}$ and ${\text{NaN}}{{\text{O}}_{\text{2}}}$ . Therefore Henderson-Hasselbalch equation becomes as follows:

${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right]}}{{\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right]}}$ …… (2)

Initial moles of ${\text{HN}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Moles of HN}}{{\text{O}}_2} &= \left( {1.2{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 1.2{\text{ mol}} \\\end{aligned}$

Initial moles of ${\text{NaN}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_{\text{2}}} &= \left( {0.8{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 0.8{\text{ mol}} \\\end{aligned}$

Moles of NaOH can be calculated as follows:

$\begin{aligned}{\text{Moles of NaOH}} &= \left( {0.5{\text{ M}}} \right)\left( {{\text{400 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.2{\text{ mol}} \\\end{aligned}$

When addition of 0.2 moles of NaOH is done to the buffer solution, 0.2 moles of ${\text{HN}}{{\text{O}}_{\text{2}}}$ is neutralized while the same amount of ${\text{NaN}}{{\text{O}}_{\text{2}}}$ is formed. Since volumes are additive, total volume can be calculated as follows:

$\begin{aligned}{\text{Total volume of solution}} &= \left( {1 + \left( {400{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)} \right){\text{ L}} \\ &= {\text{1}}{\text{.4 L}} \\\end{aligned}$

Therefore concentration of ${\text{HN}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 - 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}}\\&= 0.714{\text{ M}} \\\end{aligned}$

Therefore concentration of ${\text{NaN}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 + 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}} \\ &= 1{\text{ M}} \\\end{aligned}$

Substitute 0.714 M for $\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right]$ , 1 M for $\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right]$ and 3.4 for ${\text{p}}{K_{\text{a}}}$ in equation (2).