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.
[tex] n_{a} = n_{b} [/tex]
Where "a" is for acid and "b" is for base.
The number of moles is:
[tex] n = C*V [/tex]
Where "C" is for concentration and "V" for volume.
[tex] C_{a}V_{a} = C_{b}V_{b} [/tex]
[tex] C_{a} = \frac{0.1293 M*28.15 mL}{5.00 mL} = 0.73 M [/tex]
Hence the initial concentration of the acid is 0.73 M.
The original pH of the acid is given by:
[tex] pH = -log([H^{+}]) [/tex]
[tex] pH = -log(0.73) = 0.14 [/tex]
Therefore, the original pH of the acid is 0.14.
I hope it helps you!
