Question

A chemist dissolves 248mg of pure hydrochloric acid in enough water to make up 140mL of solution.Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The pH of the solution is, 1.31

Explanation : Given,

Mass of $HCl$ = 248 mg = 0.248 g      (1 mg = 0.001 g)

Volume of solution = 140 mL

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the concentration of hydrochloric acid.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }HCl\times 1000}{\text{Molar mass of }HCl\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.248g\times 1000}{36.5g/mole\times 140mL}=0.0485mole/L=0.0485M$

The concentration of HCl is, 0.0485 M

As we know that, HCl dissociates to give hydrogen ion and chloride ion.

The dissociation reaction is:

$HCl\rightarrow H^++Cl^-$

Concentration of HCl = Concentration of H⁺ = Concentration of Cl⁻ = 0.0485 M

Now we have to calculate the pH of the solution.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

Mathematically,

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

$pH=-\log (0.0485)$

$pH=1.31$

Therefore, the pH of the solution is, 1.31