Question

1. what is the h₃o⁺ in a solution that is 0.0515 m hno₂ and 0.120 m nano₂?
2. what is the ph of a buffer solution that contains 0.012 m benzoic acid, c₆h₅cooh (kₐ = 6.2 x 10⁻⁵), and 0.033 m sodium benzoate, nac₆h₅cooh?
3. what mass of sodium benzoate should you add to 100.0 ml of 0.15 m benzoic acid to obtain a buffer with ph = 4.20? assume no volume change. (see #2 for the ka)

Explanation:

1.

Step1: Write the Henderson - Hasselbalch equation for the $HNO_2/NaNO_2$ buffer

The Henderson - Hasselbalch equation for a buffer of a weak acid $HA$ and its conjugate base $A^-$ is $pH = pK_a+\log\frac{[A^-]}{[HA]}$, and $[H_3O^+]=10^{-pH}$. For $HNO_2$, $K_a = 4.5\times 10^{-4}$, so $pK_a=-\log(4.5\times 10^{-4})\approx 3.35$. Here, $[HA]=0.0515\ M$ and $[A^-]=0.120\ M$.

Step2: Calculate the $pH$

$pH = pK_a+\log\frac{[A^-]}{[HA]}=3.35+\log\frac{0.120}{0.0515}=3.35 + \log(2.33)=3.35+ 0.37=3.72$

Step3: Calculate $[H_3O^+]$

$[H_3O^+]=10^{-pH}=10^{-3.72}\approx 1.9\times 10^{-4}\ M$

2.

Step1: Identify the values for the Henderson - Hasselbalch equation

For benzoic acid $C_6H_5COOH$, $K_a = 6.2\times 10^{-5}$, so $pK_a=-\log(6.2\times 10^{-5})\approx 4.21$. $[HA]=0.012\ M$ and $[A^-]=0.033\ M$.

Step2: Calculate the $pH$

$pH = pK_a+\log\frac{[A^-]}{[HA]}=4.21+\log\frac{0.033}{0.012}=4.21+\log(2.75)=4.21 + 0.44=4.65$

3.

Step1: Rearrange the Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is $pH = pK_a+\log\frac{[A^-]}{[HA]}$. We know $pH = 4.20$, $pK_a = 4.21$ (from above), and $[HA]=0.15\ M$. Rearranging for $\frac{[A^-]}{[HA]}$ gives $\log\frac{[A^-]}{[HA]}=pH - pK_a$.

Step2: Calculate $\frac{[A^-]}{[HA]}$

$\log\frac{[A^-]}{[HA]}=4.20 - 4.21=- 0.01$, so $\frac{[A^-]}{[HA]}=10^{-0.01}\approx 0.98$.

Step3: Find $[A^-]$

Since $[HA]=0.15\ M$, $[A^-]=0.98\times0.15\ M = 0.147\ M$.

Step4: Calculate the moles of $A^-$

In a $100.0\ mL(0.100\ L)$ solution, the number of moles of $A^-$ (sodium benzoate) $n = [A^-]\times V=0.147\ M\times0.100\ L = 0.0147\ mol$.

Step5: Calculate the mass of sodium benzoate

The molar mass of sodium benzoate $C_6H_5COONa$ is $M=(7\times12 + 5\times1+1\times16 + 1\times23)=144\ g/mol$. The mass $m=n\times M=0.0147\ mol\times144\ g/mol\approx 2.12\ g$

Answer:

1. $[H_3O^+]\approx 1.9\times 10^{-4}\ M$
2. $pH = 4.65$
3. Mass of sodium benzoate $\approx 2.12\ g$