What is the pH of an aqueous solution prepared by dissolving 0.365\text{ g} of pure hydrog… — JAMB UTME

What is the pH of an aqueous solution prepared by dissolving $0.365\text{ g}$ of pure hydrogen chloride gas ( $\text{HCl}$ ) in distilled water to make a total solution volume of $10.0\text{ dm}^3$ at $25^\circ\text{C}$ ? [Molar mass: $\text{H} = 1.0\text{ g mol}^{-1}, \text{Cl} = 35.5\text{ g mol}^{-1}$ ]

A
1.0
B
2.0
3.0Answer
D
11.0

Answer

The pH of the resulting solution is 3.0.

Dissolving

0.365\text{ g}

\text{HCl}

(molar mass

36.5\text{ g mol}^{-1}

) yields

0.01\text{ mol}

\text{HCl}

. In a

10.0\text{ dm}^3

solution, the hydrogen ion concentration is

\frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. Taking the negative logarithm gives a pH of

3.0

Step-by-Step Solution

Calculate the molar mass of

\text{HCl}

and find the number of moles dissolved.

\text{Molar mass of HCl} = 1.0 + 35.5 = 36.5\text{ g mol}^{-1}

. Moles of

\text{HCl} = \frac{0.365\text{ g}}{36.5\text{ g mol}^{-1}} = 0.01\text{ mol} = 1.0 \times 10^{-2}\text{ mol}

pH depends on the molar concentration of hydrogen ions, which requires knowing the total moles of solute.

Determine the molarity (

[\text{H}^+]

) of the solution.

[\text{H}^+] = \frac{\text{Moles of solute}}{\text{Volume in dm}^3} = \frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

\text{HCl}

is a strong monoprotic acid and ionizes completely in water to yield equal moles of

\text{H}^+

ions.

Calculate the pH using the pH definition formula.

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

The negative logarithm of the hydrogen ion concentration gives the pH value of the solution.

Key Concept

pH Calculation of a Strong Acid from Mass and Volume

Estimated Time:1m 30s

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