A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M...

A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M NaOH. K1 = 1.49 x 10-2, K2 = 2.03 x 10-6. What is the pH after 50.00 mL of NaOH has been added?

Expert Solution

no of moles of H2A = molarity * volume in L

= 1*0.1 = 0.1moles

no of moles of NaOH = molarity * volume in L

= 1*0.05 = 0.05 moles

H2A (aq) + NaOH(aq) --------------> NaHA(aq) + H2O

I 0.1 0.05 0

C -0.05 -0.05 0.05

E 0.05 0 0.05

Pka = -logKa

= -log1.49*10^-2

= 1.8268

PH = PKa + log[NaHA]/[H2A]

= 1.8268 + log0.05/0.05

= 1.8268 + log1

= 1.8268 >>>>answer

queen_honey_blossom answered 3 years ago

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