Question

question 4 (8 points)
150.00 ml of 1.50 mol/l methanoic acid is being titrated with 0.750 mol/l sodium hydroxide. what is the ph after 30.00 ml of base has been added?
(k_a = 1.78 x 10^{-4})

ph = 4.00
ph = 7.00
ph = 0.140
ph = 2.80

Explanation:

Step1: Calculate moles of acid and base

Moles of methanoic acid ($n_{HA}$) = $M\times V=1.50\ mol/L\times0.150\ L = 0.225\ mol$
Moles of sodium hydroxide ($n_{OH^ -}$) = $M\times V = 0.750\ mol/L\times0.030\ L=0.0225\ mol$

Step2: Determine moles of acid remaining and moles of conjugate - base formed

After the reaction, moles of $HA$ remaining = $n_{HA}-n_{OH^ -}=0.225 - 0.0225=0.2025\ mol$
Moles of $A^ -$ (conjugate - base) formed = $n_{OH^ -}=0.0225\ mol$
Total volume = $150.00\ mL + 30.00\ mL=180.00\ mL = 0.180\ L$

Step3: Calculate concentrations of $HA$ and $A^ -$

$[HA]=\frac{0.2025\ mol}{0.180\ L}=1.125\ mol/L$
$[A^ -]=\frac{0.0225\ mol}{0.180\ L}=0.125\ mol/L$

Step4: Use the Henderson - Hasselbalch equation

$pH = pK_a+\log\frac{[A^ -]}{[HA]}$
$pK_a=-\log(K_a)=-\log(1.78\times 10^{-4})\approx3.75$
$pH = 3.75+\log\frac{0.125}{1.125}=3.75+\log(0.111)\approx3.75 - 0.95=2.80$

Answer:

$pH = 2.80$