Question

How much NaOH (in grams) is needed to prepare 463 mL of solution with a pH of 10.020?

Answer 1

Answer: $1.94\times 10^{-3}g$ of NaOH

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

Putting in the values:

$10.020=-\log[H^+]$

$[H^+]=9.55\times 10^{-11}$

$[H^+][OH^-]=10^{-14}$

$[OH^-]=\frac{10^{-14}}{9.55\times 10^{-11}}=1.05\times 10^{-4}M$

$NaOH\rightarrow Na^++OH^-$

$Molarity=\frac{moles\times 1000}{\text {Volume in ml}}$

$1.05\times 10^{-4}M=\frac{moles\times 1000}{463ml}$

moles = $4.86\times 10^{-5}$

Mass of $NaOH=moles\times {\text {Molar mass}}=4.86\times 10^{-5}\times 40=1.94\times 10^{-3}g$

Thus $1.94\times 10^{-3}g$ of NaOH is needed to prepare 463 mL of solution with a pH of 10.020