Title: Determining the dissociation of a weak acid using pH Measurements Data: M

Question

Title: Determining the dissociation of a weak acid using pH Measurements

Data:

Molarity of unknown acid 1.0 M

NaOH solution: 0.500 M

Volume of Unknown acid, mL: 20.0 mL

Final Buret Reading, mL: 18.2 mL

Initial Buret Reading, mL: 6.25 mL

Volume of NaOH solution, mL: 11.95 mL

Total Volume of solution, mL: .100 mL

pH reading: 4.32

I need to find the following: Calculating Ka of unknown acid

Initial number of moles for HAn(aq) & OH- (aq)

Number of moles at equilibrium: HAn(aq) & An-

Equilibrium concentration, mol L^-: HAn, An- & H3O^+

Ka ?

Please show work so I can understand it

Explanation / Answer

M1 = 0.5M

V1 = 11.95ml

M2 = 1M

V2 = 20ml

total volume = 31.95ml = 0.3195L

pH = -log10[H+]

[H+] = 4.786*10^-5 M

moles = 4.786*10^-5 * 0.1 = 4.786*10^-6 moles

moles of H+ consumed by NaOH = 0.5*0.01195 = 5.975*10^-3 moles

initial moles of H+ = 6.023*10^-3

initial moles of acid = 0.02*1 = 0.02moles

let the acid be HA

HA ---> H+ + A-

at eq 0.02-x x x

and x = 6.023*10^-3

so keq = [H+][A-]/[HA]

keq = x^2/(0.02-x) = 2.595*10^-3..................ANS

initial moles of HA= 0.02moles

initial moles of OH- = 5.975*10^-3 moles

moles of HA at eq = 0.02-6.023*10^-3 = 0.0139 moles

moles of A- at eq = x = 6.023*10^-3 moles

equilibrium concentration

[HA] = 0.0139/0.1 = 0.139 M

[A-] = 6.023*10^-3 / 0.1= 6.023*10^-2M

[H3O+] = 4.786*10^-5 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.