I am having problems with my chemistry lab homework. Title: Determining the diss

Question

I am having problems with my chemistry lab homework. Title: Determining the dissociation of a weak acid using pH Measurements

Data:

Molarity of unknown acid1.0 M

NaOH solution: 0.500 M

Volume of Unknown acid, mL: 20.0 mL

Final Buret Reading, mL: 6.25 mL

Initial Buret Reading, mL: 0.00 mL

Volume of NaOH solution, mL: 6.25 mL

Total Volume of solution, mL: .100 mL

pH reading: 4.10

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 ?

Slope of the line ?

Ka Determined slope

Please help

Plz show work so I can understand it

Explanation / Answer

M1 = 0.5M

V1 = 6.25ml

M2 = 1M

V2 = 20ml

total volume = 100ml = 0.1L

pH = -log10[H+]

[H+] = 7.94*10^-4 M

moles = 7.94*10^-4 * 0.1 = 7.94*10^-5 moles

moles of H+ consumed by NaOH = 0.5*0.00625 = 0.003125 moles

initial moles of H+ present = 3.2044*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 = 3.2044*10^-3

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

keq = x^2/(0.02-x) = 6.11*10^-4...................................ANS

initial moles of HA= 0.02moles

initial moles of OH- = 0.003125 moles

moles of HA at eq = 0.02-3.2044*10^-3 = 0.01679 moles

moles of A- at eq = x = 3.2044*10^-3 moles

equilibrium concentration

[HA] = 0.01679/0.1 = 0.1679 M

[A-] = 3.2044*10^-3 / 0.1= 3.2044*10^-2M

[H3O+] = 7.94*10^-4 M

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