Question

1. determine the ph of the acid solution after the following additions have been made during the titration. the initial volume of each acid is 30.00 ml.

addition of 0.300 m naoh	0.300 m hno₃	0.300 m hno₂

|0 ml|
|10.00 ml|
|15.00 ml|
|30.00 ml|
|35.00 ml|

Explanation:

Step1: Analyze $HNO_3$ (strong - acid) cases

$HNO_3$ is a strong acid, so $[H^+]=[HNO_3]$. For initial case ($V_{NaOH}=0$ mL), $n_{HNO_3}=M\times V = 0.300\ M\times0.0300\ L = 0.00900\ mol$, $[H^+]=0.300\ M$, $pH = -\log[H^+]=-\log(0.300)\approx0.52$.
When $V_{NaOH} = 10.00\ mL$, $n_{NaOH}=0.300\ M\times0.0100\ L = 0.00300\ mol$, $n_{HNO_3\ remaining}=0.00900 - 0.00300=0.00600\ mol$, total volume $V = 0.0300\ L+0.0100\ L = 0.0400\ L$, $[H^+]=\frac{0.00600\ mol}{0.0400\ L}=0.150\ M$, $pH=-\log(0.150)\approx0.82$.
When $V_{NaOH}=15.00\ mL$, $n_{NaOH}=0.300\ M\times0.0150\ L = 0.00450\ mol$, $n_{HNO_3\ remaining}=0.00900 - 0.00450 = 0.00450\ mol$, $V = 0.0300\ L + 0.0150\ L=0.0450\ L$, $[H^+]=\frac{0.00450\ mol}{0.0450\ L}=0.100\ M$, $pH = -\log(0.100)=1.00$.
When $V_{NaOH}=30.00\ mL$, $n_{NaOH}=0.300\ M\times0.0300\ L = 0.00900\ mol$, it completely neutralizes $HNO_3$, $pH = 7.00$.
When $V_{NaOH}=35.00\ mL$, $n_{NaOH}=0.300\ M\times0.0350\ L=0.0105\ mol$, $n_{OH^-}=0.0105 - 0.00900 = 0.00150\ mol$, $V = 0.0300\ L+0.0350\ L = 0.0650\ L$, $[OH^-]=\frac{0.00150\ mol}{0.0650\ L}\approx0.0231\ M$, $pOH=-\log(0.0231)\approx1.64$, $pH = 14 - pOH=12.36$.

Step2: Analyze $HNO_2$ (weak - acid) cases

$HNO_2$ has $K_a = 4.5\times10^{-4}$. Initial case ($V_{NaOH}=0$ mL), let $[H^+]=x$, $K_a=\frac{x\cdot x}{0.300 - x}\approx\frac{x^2}{0.300}$ (since $0.300 - x\approx0.300$ as $K_a$ is small), $x=\sqrt{K_a\times0.300}=\sqrt{4.5\times10^{-4}\times0.300}\approx1.16\times10^{-2}\ M$, $pH=-\log(1.16\times10^{-2})\approx1.94$.
When $V_{NaOH}=10.00\ mL$, $n_{NaOH}=0.300\ M\times0.0100\ L = 0.00300\ mol$, $n_{HNO_2\ remaining}=0.00900 - 0.00300 = 0.00600\ mol$, $n_{NO_2^-}=0.00300\ mol$, using the Henderson - Hasselbalch equation $pH=pK_a+\log\frac{[A^-]}{[HA]}$, $pK_a=-\log(4.5\times10^{-4})\approx3.35$, $pH = 3.35+\log\frac{0.00300/0.0400}{0.00600/0.0400}=3.05$.
When $V_{NaOH}=15.00\ mL$, $n_{NaOH}=0.300\ M\times0.0150\ L = 0.00450\ mol$, $n_{HNO_2\ remaining}=0.00900 - 0.00450 = 0.00450\ mol$, $n_{NO_2^-}=0.00450\ mol$, $pH=pK_a+\log\frac{0.00450/0.0450}{0.00450/0.0450}=3.35$.
When $V_{NaOH}=30.00\ mL$, it forms a $NO_2^-$ solution. $n_{NO_2^-}=0.00900\ mol$, $V = 0.0600\ L$, $[NO_2^-]=0.150\ M$. $K_b=\frac{K_w}{K_a}=\frac{1.0\times10^{-14}}{4.5\times10^{-4}}\approx2.22\times10^{-11}$, let $[OH^-]=y$, $K_b=\frac{y\cdot y}{0.150 - y}\approx\frac{y^2}{0.150}$, $y=\sqrt{K_b\times0.150}=\sqrt{2.22\times10^{-11}\times0.150}\approx1.83\times10^{-6}\ M$, $pOH = -\log(1.83\times10^{-6})\approx5.74$, $pH=14 - 5.74 = 8.26$.
When $V_{NaOH}=35.00\ mL$, $n_{NaOH}=0.300\ M\times0.0350\ L = 0.0105\ mol$, $n_{OH^-}=0.0105 - 0.00900=0.00150\ mol$, $V = 0.0300\ L+0.0350\ L = 0.0650\ L$, $[OH^-]=\frac{0.00150\ mol}{0.0650\ L}\approx0.0231\ M$, $pOH=-\log(0.0231)\approx1.64$, $pH = 12.36$.

Answer:

Addition of 0.300 M NaOH	0.300 M $HNO_3$	0.300 M $HNO_2$
10.00 mL	$\approx0.82$	$\approx3.05$
15.00 mL	1.00	3.35
30.00 mL	7.00	8.26
35.00 mL	12.36	12.36