Question

What is the pH of a solution containing 0.35 M phenol (Ka = 1.3 x 10-10) and 0.40 M sodium phenoate?

Answer 1

Answer on the Question #68539, Chemistry / Other

What is the pH of a solution containing 0.35 M phenol (Ka = 1.3 x 10⁻¹⁰) and 0.40 M sodium phenoate?

Solution:

pH is the negative logarithm of equilibrium concentration of Hydrogen ions [H⁺]:

$$pH = -\lg[H^{+}]$$

To calculate equilibrium concentration of [H⁺] using following equation:

$$[H^{+}] = K_{a} \frac{C(C_{6}H_{5}OH)}{C(C_{6}H_{5}ONa)} = 1.3 \cdot 10^{-10} \frac{0.35M}{0.40M} = 1.1 \cdot 10^{-10}$$

After we can calculate pH of buffer solution:

$$pH = -\lg[H^{+}] = -\lg(1.1 \cdot 10^{-10}) = 9.96$$

**Answer**: pH of solution is 9.96

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