What is the pH of the solution created when 1.00 mL of NaOH is titrated into...

Question

Question

What is the pH of the solution created when 1.00 mL of NaOH is titrated into 10 mL of 0.1 M acetic acid and acetate buffer solution?

Concentration of NaOH is 0.220849 M.

Answer 1

Answer #1

no of moles of CH3COOH = molarity * volume in L

= 0.1*0.01 = 0.001moles

no of moles of NaOH = molarity * volume in L

= 0.220849*0.001 = 0.000220849moles

PKa = 4.75

------------ CH3COOH(aq) + NaOH(aq) -------------> CH3COONa(aq) + H2O(l)

I------------- 0.001---------------------- 0.000220849 ------------------ 0

C----------- -0.000220849 ---------- -0.000220849-------------------0.000220849

E------------ 0.000779151-------------- 0 -------------------------------0.000220849

PH = PKa + log[CH3COONa]/[CH3COOH]

= 4.75 + log0.000220849/0.000779151

= 4.75 -0.5475

= 4.2025 >>>>>answer

Answer 2

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