Question

A 5.00 mL sample of hydrochloric acid is titrated with 0.1293 M ammonia (a base). If the titration required 28.15 mL of ammonia, determine the following:
the original concentration of the acid
the original pH of the acid

Answer 1

Answer:

1. C = 0.73 M.

2. pH = 0.14

       

Explanation:

The reaction is the following:

HCl + NH₃ ⇄ NH₄⁺Cl⁻

From the titration, we can find the number of moles of HCl that were neutralized by the ammonia.

$n_{a} = n_{b}$

Where "a" is for acid and "b" is for base.

The number of moles is:

$n = C*V$  

Where "C" is for concentration and "V" for volume.

$C_{a}V_{a} = C_{b}V_{b}$

$C_{a} = \frac{0.1293 M*28.15 mL}{5.00 mL} = 0.73 M$

Hence the initial concentration of the acid is 0.73 M.

The original pH of the acid is given by:

$pH = -log([H^{+}])$

$pH = -log(0.73) = 0.14$          

Therefore, the original pH of the acid is 0.14.

I hope it helps you!