What is the pH of the NaOH solution when 0.04 | vedclass

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Question

(D) The molecular mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$. The number of moles of $NaOH = \frac{0.04 \ g}{40 \ g/mol} = 0.001 \ mol = 10^{-3} \ mol$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The molecular mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$. The number of moles of $NaOH = \frac{0.04 \ g}{40 \ g/mol} = 0.001 \ mol = 10^{-3} \ mol$. The volume of the solution = $100 \ mL = 0.1 \ L$. Molarity of $NaOH$ solution = $\frac{10^{-3} \ mol}{0.1 \ L} = 10^{-2} \ M$. Since $NaOH$ is a strong base,$[OH^-] = 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. Using the relation $pH + pOH = 14$,we get $pH = 14 - 2 = 12$.

What is the $pH$ of the $NaOH$ solution when $0.04 \ g$ of it is dissolved in water to make a $100 \ mL$ solution?

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