Question

Calculate the pH of a Sr(OH)2 solution prepared by dissolving 0.3211 g of Sr(OH)2 and making up to 250.00 mL solution

Answer 1

Moles of Sr(OH)₂ in 250ml.

$\frac{mass}{molar\space mass}=\frac{0.3211}{121.63}=0.0026399736$

We can use the following relation to find molarity.

0.0026399736 moles $\equiv$ 250 ml

For 1000 ml we have

$\frac{1000}{250}×0.0026399736=0.01056 \space moles$

pOH=-log [Sr (OH)₂]

pH for bases is essentially 14-pOH.

$pH=14-(-log\space0.01056)=12.024$