Question

The equation for the reaction of strontium hydroxide with hydrochloric acid is shown below.

$Sr {(OH)}_{2} + 2 HCl \to {SrCl}_{2} + 2 H_{2} O$

$25.0 {cm}^{3}$ of a solution of strontium hydroxide was exactly neutralised by $15.00 {cm}^{3}$ of $0.100 mol {dm}^{- 3}$ hydrochloric acid. Calculate the concentration, in $mol {dm}^{- 3}$ , of the strontium hydroxide solution.

Accepted Answer

First we have to calculate the moles number of hydrochloric acid

Volume of hydrochloric acid = $15.00 {cm}^{3} = \frac{15.00}{1000} = 0.015 {dm}^{3}$

(It is needed to convert all the units into one type of measuring system)

$Number of moles of solute = Molarity \times Volume of solvent$

$= 0.100 \times 0.015 = 0.0015 moles$

From the moles of hydrochloric acid we have to calculate how many moles of strontium hydroxide reacted by using stoichiometric equation,

$Sr {(OH)}_{2} + 2 HCl \to {SrCl}_{2} + 2 H_{2} O$

From the above stoichiometric equation,

Moles of strontium hydroxide $=$ Moles of hydrochloric acid $\times \frac{1}{2}$

$= 0.0015 \times \frac{1}{2} = 0.00075 moles$

Then the concentration of strontium hydroxide $= ?$

We have given,

Volume of strontium hydroxide $= 25.0 {cm}^{3} = \frac{25.0}{1000} = 0.025 {dm}^{3}$

$Molarity = \frac{Number of moles of solute}{Voume of solution ({dm}^{3})}$

$= \frac{0.00075}{0.025} = 0.03 mol {dm}^{- 3}$

Therefore, concentration of strontium hydroxide in $mol {dm}^{- 3}$ is $0.03 mol {dm}^{- 3}$ .

Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 23: Question

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 23: Question

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 23: Question with Hints & Solutions

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