A solution is prepared by dissolving 0.04\text{ g} of sodium hydroxide (\text{NaOH}) in di… — JAMB UTME

A solution is prepared by dissolving $0.04\text{ g}$ of sodium hydroxide ( $\text{NaOH}$ ) in distilled water to make $1.0\text{ dm}^3$ of solution at $25^\circ\text{C}$ . What is the $\text{pH}$ of the resulting solution? [Molar mass of $\text{NaOH} = 40\text{ g mol}^{-1}$ ]

A
$3.0$
$11.0$ Answer
C
$1.4$
D
$12.6$

Answer

The pH of the resulting solution is

11.0

The solution contains

0.04\text{ g}

\text{NaOH}

1.0\text{ dm}^3

, which corresponds to

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

1.0 \times 10^{-3}\text{ mol dm}^{-3}

. Since

\text{NaOH}

is a strong base,

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

, giving a

\text{pOH}

3.0

. Using the relationship

\text{pH} + \text{pOH} = 14

, the

\text{pH}

14.0 - 3.0 = 11.0

Step-by-Step Solution

Calculate the molar concentration of sodium hydroxide (

\text{NaOH}

\text{Molar concentration} = \frac{0.04\text{ g dm}^{-3}}{40\text{ g mol}^{-1}} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

Concentration must be in

\text{mol dm}^{-3}

before calculating ion concentration and

\text{pOH}

Determine the hydroxide ion concentration

[\text{OH}^-]

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

\text{NaOH}

is a strong monobasic base that completely dissociates into

\text{Na}^+

and

\text{OH}^-

ions.

Calculate the

\text{pOH}

of the solution.

\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

\text{pOH}

is defined as the negative base-10 logarithm of the hydroxide ion concentration.

Convert

\text{pOH}

\text{pH}

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

25^\circ\text{C}

\text{pH} + \text{pOH} = 14.0

Key Concept

Calculating pH from the mass concentration of a strong base via pOH conversion

Estimated Time:1m 30s

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