Chemistry, 23.06.2019 03:40 ElegantEmerald
The following questions a24 - a26 relate to 100 ml of 0.0150 m solution of benzoic acid (c6h3cooh). ka(c6h3cooh) = 6.4 x 10^-5. what is the ph of the solution after the addition of 1 x 10^-3 moles of naoh? you may assume no volume change to the solution upon addition of the naoh.
Answers: 2
be 
. Note that 
.
![\displaystyle \frac{[\mathrm{C_6H_5COO^{-}}]\cdot [\mathrm{H^{+}}]}{[\mathrm{C_6H_5COOH}]} = \mathrm{pK}_{a}](/tpl/images/0006/5046/265ec.png)
.
.
:
.
.
.![\rm [H^{+}] = 0.00306655\; mol\cdot L^{-1}](/tpl/images/0006/5046/db736.png)
.![\displaystyle \mathrm{pH} = -\log_{10}{[\mathrm{H^{+}}]} = 2.513](/tpl/images/0006/5046/141cf.png)
.
. Benzoic acid is thus a monoprotic acid. Each mole of the acid will react with only one mole of 
. The 100 mL solution initially contains 
moles of benzoic acid. The 
moles of 
(from the salt 
) and 
moles of 
.
and the conjugate base of the acid 
for benzoic acid.![\begin{aligned}\mathrm{pH} &= \mathrm{pK}_{a} + \log{\frac{{[\text{Conjugate Base}]}}{[\text{Weak Acid}]}} \\ &= \mathrm{pK}_{a} + \log{\frac{{[\mathrm{C_6H_5COO^{-}}]}}{[\mathrm{C_6H_5COOH}]}}\\ &= 4.19382 + \log{\frac{0.01}{0.005}}\\ &\approx 4.495 \end{aligned}](/tpl/images/0006/5046/20b2b.png)
.
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