Question

pH values of 0.1 M HCL and 0.1 M CH3COOH are

Answer 1

Answer on Question #41127 - Chemistry - Other

Question:

pH values of 0.1 M HCl and 0.1 M CH₃COOH are

Answer:

1. pH equals:

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

HCl is a strong acid and it fully dissociates in water:

$$\mathrm{HCl}_{(\mathrm{aq})} \leftrightarrow \mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{Cl}^{-}_{(\mathrm{aq})}$$

We see that concentration of HCl equals $\mathrm{H}^{+}$ ion concentration. Therefore $\mathrm{H}^{+}$ ion concentration is:

$$\left[ \mathrm{H}^{+} \right] = \mathrm{C}(\mathrm{HCl}) = 0.1\ \mathrm{M} = 0.1\ \mathrm{mol/L}$$

pH value of 0.1 M HCl is:

$$\mathrm{pH} = -\lg (0.1) = 1$$

2. Acetic acid $\mathrm{CH_3COOH}$ ionizes partially, so we need the pKa or Ka value of acetic acid. Dissociation of $\mathrm{CH_3COOH}$:

$$\mathrm{CH_3COOH}_{(\mathrm{aq})} \leftrightarrow \mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{CH_3COO}^{-}_{(\mathrm{aq})}$$

Ka is dissociation constant which is concentration of product/concentration of reactant:

$$\mathrm{Ka} = \left[ \mathrm{CH_3COO}^{-} \right] * \left[ \mathrm{H}^{+} \right] / \left[ \mathrm{CH_3COOH} \right]$$

After partial dissociation, $\left[ \mathrm{CH_3COOH} \right]$ is still approximately 0.1 M, and $\left[ \mathrm{CH_3COO}^{-} \right] = \left[ \mathrm{H}^{+} \right]$.

Ka of $\mathrm{CH_3COOH} = 1.74 * 10^{-5}$ (from data book). Then $\mathrm{H}^{+}$ ion concentration equals:

$$\begin{array}{l} 1.74 * 10^{-5} = \left[ \mathrm{H}^{+} \right]^{2} / 0.1 \\ \left[ \mathrm{H}^{+} \right]^{2} = 1.74 * 10^{-5} * 0.1 \\ \left[ \mathrm{H}^{+} \right]^{2} = 1.74 * 10^{-6} \\ \left[ \mathrm{H}^{+} \right] = 1.32 * 10^{-3} \\ \end{array}$$

pH value of 0.1 M $\mathrm{CH_3COOH}$ is:

$$\mathrm{pH} = -\lg (1.32 * 10^{-3}) = 2.88$$

Answer:

1. pH = 1

2. pH = 2.88

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