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

1. C = 0.73 M.

2. pH = 0.14

Step-by-step 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) = (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!