Question

What is the pH of the solution when 10.005g of sodium hydroxide is dissolved in 500mL of 0.5M acetic acid. pKa value for acetic acid is 4.74.

Answer 1

we have below equation to be used:

pKa = -log Ka

4.74 = -log Ka

log Ka = -4.74

Ka = 10^(-4.74)

Ka = 1.82*10^-5

we have:

Molarity of HC2H3O2 = 0.5 M

Volume of HC2H3O2 = 0.5 L

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH = 10.005 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(10.005 g)/(39.998 g/mol)

= 0.2501 mol

mol of HC2H3O2 = Molarity of HC2H3O2 * Volume of HC2H3O2

mol of HC2H3O2 = 0.5 M * 0.5 L = 0.25 mol

We have:

mol of HC2H3O2 = 0.25 mol

mol of NaOH = 0.2501 mol

0.2501 mol of both will react to form C2H3O2- and H2O

C2H3O2- here is strong base

C2H3O2- formed = 0.25 mol

Volume of Solution = 0.5 + 0 = 0.5 L

Kb of C2H3O2- = Kw/Ka = 1*10^-14/1.82*10^-5 = 5.495*10^-10

concentration ofC2H3O2-,c = 0.25 mol/0.5 L = 0.5M

C2H3O2- dissociates as

C2H3O2- + H2O -----> HC2H3O2 + OH-

0.5 0 0

0.5-x x x

Kb = [HC2H3O2][OH-]/[C2H3O2-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.495*10^-10)*0.5) = 1.657*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.657*10^-5 M

[OH-] = x = 1.657*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.657*10^-5)

= 4.78

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.78

= 9.22

Answer: 9.22