find the ph during the titration of 20.00 ml of 0.1000 m butanoic acid, ch₃ch₂ch₂cooh, with 0.1000 m naoh…

find the ph during the titration of 20.00 ml of 0.1000 m butanoic acid, ch₃ch₂ch₂cooh, with 0.1000 m naoh solution after the following additions of titrant.\nnote: reference the (k_{a}) of acids at 25 °c table for additional information.\npart 1 of 3\nround your answer to 3 decimal places.\n13.00 ml:\nph = \npart 2 of 3\nround your answer to 2 decimal places.\n20.10 ml:\nph =

find the ph during the titration of 20.00 ml of 0.1000 m butanoic acid, ch₃ch₂ch₂cooh, with 0.1000 m naoh solution after the following additions of titrant.\nnote: reference the (k_{a}) of acids at 25 °c table for additional information.\npart 1 of 3\nround your answer to 3 decimal places.\n13.00 ml:\nph = \npart 2 of 3\nround your answer to 2 decimal places.\n20.10 ml:\nph =

Answer

Explanation:

Step1: Calculate moles of acid and base

Moles of butanoic acid ($n_{HA}$) initially = $M\times V = 0.1000\ M\times20.00\times10^{- 3}\ L=2.000\times10^{-3}\ mol$. Moles of NaOH added ($n_{OH^ -}$) for 13.00 mL addition = $0.1000\ M\times13.00\times10^{-3}\ L = 1.300\times10^{-3}\ mol$.

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

Remaining moles of butanoic acid ($n_{HA_{remaining}}$) = $2.000\times10^{-3}-1.300\times10^{-3}=7.000\times10^{-4}\ mol$. Moles of butanoate ion ($n_{A^ -}$) formed = $1.300\times10^{-3}\ mol$. Total volume ($V_{total}$) = $(20.00 + 13.00)\ mL=33.00\ mL = 0.033\ L$. Concentration of butanoic acid ($[HA]$) = $\frac{7.000\times10^{-4}\ mol}{0.033\ L}\approx0.0212\ M$. Concentration of butanoate ion ($[A^ -]$) = $\frac{1.300\times10^{-3}\ mol}{0.033\ L}\approx0.0394\ M$. The $K_a$ of butanoic acid ($CH_3CH_2CH_2COOH$) is $1.52\times10^{-5}$ (from reference table).

Step3: Use the Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is $pH = pK_a+\log\frac{[A^ -]}{[HA]}$. $pK_a=-\log(K_a)=-\log(1.52\times10^{-5})\approx4.818$. $pH = 4.818+\log\frac{0.0394}{0.0212}=4.818+\log(1.86)\approx4.818 + 0.269=5.087$.

For the addition of 20.10 mL of NaOH:

Step1: Calculate moles of acid and base

Moles of NaOH added ($n_{OH^ -}$) = $0.1000\ M\times20.10\times10^{-3}\ L=2.010\times10^{-3}\ mol$. Moles of butanoic acid ($n_{HA}$) initially = $0.1000\ M\times20.00\times10^{-3}\ L = 2.000\times10^{-3}\ mol$.

Step2: Determine excess moles of base

Excess moles of NaOH = $2.010\times10^{-3}-2.000\times10^{-3}=1.0\times10^{-5}\ mol$. Total volume ($V_{total}$) = $(20.00 + 20.10)\ mL = 40.10\ mL=0.0401\ L$. Concentration of excess $OH^ -$ ($[OH^ -]$) = $\frac{1.0\times10^{-5}\ mol}{0.0401\ L}\approx2.494\times10^{-4}\ M$.

Step3: Calculate pOH and then pH

$pOH=-\log([OH^ -])=-\log(2.494\times10^{-4})\approx3.603$. $pH = 14 - pOH=14 - 3.603 = 10.397\approx10.40$.

Answer:

Part 1 of 3: 5.087 Part 2 of 3: 10.40