Calculate the pH?

Question

Consider the titration of 30.0 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.

(a) 0 mL

(b) 10.0 mL

(c) 20.0 mL

Answer 1

NH3 + HCl => NH4Cl

the moles of NH3 : 0.03 * 0.003 = 9*10^-5 < 0.025
thus, NH3 will react completely
the moles of HCl left : 0.025 - 9*10^-5 = 0.02491 moles
the moles of H+ = the moles of HCl left

(a) the molar concentration of H+ : 0.002491 / 0.03 = 0.83
pH = -log[H+] = 0.08

(b) the molar concentration of H+ : 0.002491 / 0.04 = 0.62275
pH = -log[H+] = 0.206

(c) the molar concentration of H+ : 0.002491 / 0.05 = 0.4982
pH = -log[H+] = 0.303

Answer 2

titrant is HCL and you have to find out its volume in every variable in a,b and c.

the known formula of titration is: C1xV1=C2xV2 (C1= concentration of concentrated, V1=volume of concentrated)(C2=concentration of dilute, V2= volume of dilute)

calculating the pH:
pH= -log [H+]
calculating the H+ concentration:
H+= 10^ -pH

hope u benefit