What is the pH of an aqueous solution prepared by dissolving 0.49\text{ g} of tetraoxosulf… — JAMB UTME

What is the pH of an aqueous solution prepared by dissolving $0.49\text{ g}$ of tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) in distilled water to make $500\text{ cm}^3$ of solution? [Molar mass of $\text{H}_2\text{SO}_4 = 98\text{ g mol}^{-1}$ , $\log_{10} 2 = 0.301$ ]

$1.70$ Answer
B
$2.00$
C
$2.30$
D
$12.30$

Answer

The pH of the solution is

1.70

Dissolving

0.49\text{ g}

\text{H}_2\text{SO}_4

(molar mass

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

) in

0.5\text{ dm}^3

yields a

0.01\text{ mol dm}^{-3}

solution. Because tetraoxosulfate(VI) acid fully ionizes into

2\text{H}^+

and

\text{SO}_4^{2-}

, the hydrogen ion concentration

[\text{H}^+]

0.02\text{ mol dm}^{-3}

. Taking the negative logarithm base 10 gives

\text{pH} = -\log_{10}(0.02) = 1.70

Step-by-Step Solution

Calculate the amount of

\text{H}_2\text{SO}_4

in moles

\text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.49\text{ g}}{98\text{ g mol}^{-1}} = 0.005\text{ mol}

Converting mass of solute to moles using molar mass.

Determine the molar concentration of the acid solution

\text{Volume} = 500\text{ cm}^3 = 0.5\text{ dm}^3

;

\text{Molarity} = \frac{0.005\text{ mol}}{0.5\text{ dm}^3} = 0.01\text{ mol dm}^{-3}

Molarity is defined as moles of solute per cubic decimetre of solution.

Calculate the hydrogen ion concentration

[\text{H}^+]

[\text{H}^+] = 2 \times 0.01\text{ mol dm}^{-3} = 0.02\text{ mol dm}^{-3} = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

\text{H}_2\text{SO}_4

is a strong dibasic acid that ionizes completely to produce two

\text{H}^+

ions per molecule.

Calculate the pH of the solution

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.0 \times 10^{-2}) = 2 - \log_{10} 2 = 2 - 0.301 = 1.699 \approx 1.70

Applying the logarithmic definition of pH.

Key Concept

pH Calculation of Dibasic Strong Acids

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