Question

In the titration of a 25.00 mL of 0.245 M weak base (Kb= 1.0 *10^-4) being titrated by 0.365 M HCl determine the pH at equivalence point and the pH after 22.4 mL of HCl has been reached. Please explain! I'm having a hard time with these type of questions.

Answer 1

let us assume , weak base = CH3NH2

[CH3NH2 ] = 0.245 M x 0.025 L = 0.006125 mol

[HCl] = 0.0365 M x 0.0224 L = 0.0008176 mol

CH3NH2 + HCl -------------------> CH3NH3+Cl-

0.006125 mol 0.0008176 mol 0

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0.006125 -0.0008176 0 0.0008176 mol

= 0.0053 mol   

Hence,

[CH3NH2] = 0.0053 mol

[CH3NH3+] =   0.0008176 mol

Then,

pOH = -logKb + log [CH3NH3+]/ [CH3NH2]

= - log(1.0x 10^-4) + log (0.0008176)/(0.0053)

= 3.18

Then,

pH = 14 -pOH = 14 -3.18 = 10.82

Therefore,

pH = 10.82