Please show how 500 mL of 010 M NHj is titrated by 020 M per

Please show how?

50.0 mL of 0.10 M NHj is titrated by 0.20 M perchloric acid (HCIO4, a strong acid). Calculate the pH of the solution after the addition of 30.0 mL of the 0.20 M perchloric acid. At 25°C K, for NH3 = 1.8 . 10-5 and Ka for NH4+-5.6-10-10. 16. a)3.80 b) 1.90 12.10 d9,25

Solution

50.0 ml of 0.10 M NH3 = 0.050 * 0.10 = 0.0050 mole.

30.0 ml of 0.20 M perchloric acid = 0.030 * 0.20 = 0.0060 mole.

so there are excess perchloric acid.

NH3 is weak base and perchloric acid is strong acid.

excess perchloric acid = 0.060 - 0.0050 = 0.0010 mole.

total volume of solution = 50 + 30 = 80 ml.

so concentration of excess perchloric acid = 0.001 * 1000 / 80 = 0.0125 M

PH = -log [HClO4] = - log [H+] = - log (0.0125) = 1.90

Therefore option B is the answer.