2. tsm ur 3o- 14 9, 033

## Expert Answer

so I got your mistake you are not considering here that initially the ph of the solution is 10 .

so if we consider this then

Before adding hcl solution 10 = 9.25 + log [NH3]/[NH4+]

[NH3]/[NH4+] = 5.6234 -(1)

Also moles of (NH3 + NH4+) = 1M x 75 ml = 75 mmoles -(2)

=> on solving equation 1 and equation2 we get

moles of NH3 were 63.676533 mmoles

moles of NH_{4}+ were 11.323466 mmoles

after adding the HCL solution , 45 mmoles of H+ will react with NH3 to produce more NH4

according to equation

NH_{3} + H^{+} ————> NH_{4}^{ } moles of NH3 will react with H+ to produce more NH4+

hence final moles of NH3 will be 63.676533 – 45 = 18.676533 mmoles

also final moles of NH4+ will be 11.323466 + 45 = 56.3234 mmoles

now you should apply the equation which you have applied in your solution

ph = 9.25 + log [18.676533]/[56.3234] = 9.25 – 0.48 = 8.77 .