# Question #2d18a

Jan 16, 2017

The balanced equation of reaction of $C a C {O}_{3}$ with $H C l$ is as follows

$C a C {O}_{3} + 2 H C l \to C {O}_{2} + {H}_{2} O + C a C {l}_{2}$

Molar mass of $C a C {O}_{3} = \left(40 + 12 + 3 \cdot 16\right) g \text{/mol"=100g"/mol}$

So 1g $C a C {O}_{3} = \frac{1}{100} m o l = 0.01 m o l$
Let 2000cc or $2 L \text{ } H C l$ solution contains $x m o l \text{ } H C l$

As per equation the reacting ratio of $C a C {O}_{3}$ and $H C l$ is $1 : 2$
Hence $\frac{0.01}{x} = \frac{1}{2}$

So $x = 0.02 m o l$ (Assuming the acid has been completely nutralized by 1g $C a C {O}_{3}$)

Hence the strength of the $H C l$ solution is $c = \frac{0.02 m o l}{2 L} = 0.01 M$

So pH if the solution is

$= - \log c = - \log \left(0.01\right) = 2$