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: 20.00mL

Final Buret Reading, mL: 37.2 mL

Initial Buret Reading, mL: 18.76 mL

Volume of NaOH solution, mL: 18.44 mL

Total Volume of solution, mL: .100 mL

pH reading: 4.64

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 ?

Explanation / Answer

M1 = 0.5M

V1 = 18.44ml

M2 = 1M

V2 = 20ml

total volume = 38.44ml = 0.03844L

pH = -log10[H+]

[H+] = 2.291*10^-5 M

moles = 2.291*10^-5 * 0.1 = 2.291*10^-6 moles

moles of H+ consumed by NaOH = 0.5*0.01844 = 9.22*10^-3 moles

initial moles of H+ = 9.22*10^-3 + 2.291*10^-5 = 9.243*10^-3 moles

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 = 9.243*10^-3

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

keq = x^2/(0.02-x) = 7.94*10^-3..................ANS

initial moles of HA= 0.02moles

initial moles of OH- = 9.22*10^-3 moles

moles of HA at eq = 0.02-9.243*10^-3 = 0.0107 moles

moles of A- at eq = x = 9.243*10^-3 moles

equilibrium concentration

[HA] = 0.0107/0.1 = 0.107 M

[A-] = 9.243*10^-3 / 0.1= 9.243*10^-2M

[H3O+] = 2.291*10^-5 M

ask if you have any doubts

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