We’re being asked **to calculate the pH** of **a) ****0.150 M benzoic acid solution** and **b) the resulting buffer solution after adding 1.44 g sodium benzoate Na ^{+}C_{6}H_{5}COO^{- }**to 100 mL of 0.150 M benzoic acid solution.

Remember that ** weak acids** partially dissociate in water and that

The dissociation of C_{6}H_{5}CO_{2}H is as follows:

C_{6}H_{5}CO_{2}H (aq) + H_{2}O (l) ⇌ H_{3}O^{+ }(aq) + C_{6}H_{5}CO_{2}^{- }(aq)

**For this problem, we need to do these steps:**

**Part a:**

**Step 1****: ***Set up an ICE table to calculate the H*_{3}*O*^{+}* ion concentration.*

*Step 2: **Write an expression for the K*_{a}* to solve for concentration of H_{3}O^{+}*

*Step 3: **Calculate the pH from H*_{3}*O*^{+}* ion concentration*

**Part b:**

*Step 1**: Calculate the moles of each species in the buffer solution **Step 2**: **Calculate the pH of the resulting **buffer solution using **the Henderson-Hasselbalch Equation:*

$\overline{){\mathbf{p}}{\mathbf{H}}{\mathbf{=}}{\mathbf{p}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{l}}{\mathbf{o}}{\mathbf{g}}\left(\frac{\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{j}\mathbf{u}\mathbf{g}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}{\mathbf{w}\mathbf{e}\mathbf{a}\mathbf{k}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{i}\mathbf{d}}\right)}$

Benzoic acid C_{6}H_{5}COOH is a weak acid with *K *_{a }= 6.3 x 10^{-5}

C_{6}H_{5}CO_{2}H (aq) + H_{2}O (l) ⇌ H_{3}O^{+ }(aq) + C_{6}H_{5}CO_{2}^{- }(aq)

a) Calculate the pH of a 0.150 M benzoic acid solution. Show all calulations.

b) Suppose 1.44g of sodium benzoate, Na^{+}C_{6}H_{5}COO^{-}, is added to 100.0 mL of the 0.150 M benzioic acid solution. Calculate the pH of the resulting buffer solution assuming the volume of the solution does not change on addition of the solid.

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