Question

Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 760. mL of each the following solutions. (See the appendix.)

(b) 0.10 M NH₄Cl

(c) 0.10 M NH₃


(d) a solution that is 0.10 M in each NH₄⁺ and NH₃

Answer 1

b.

moles NH4+ = 0.76 L x 0.10 M=0.076

moles NH3 = 0.076 - 0.0708 = 0.0052

moles NaOH = 0.012 * 5.9 = 0.0708

0.0052

moles NH3 = 0.0052

[H+]*(0.0052/0.0708) = 10^(-9.244)

[H+] = 7.76 *10^-9

pH = 9 - log 7.76 = 8.11

c.

moles NH3 = 0.76 L x 0.10 M=0.076
moles NaOH = 0.012 * 5.9 = 0.0708

[H+]*(0.076/0.0708) = 10^(-9.244)

[H+] = 5.31 * 10^-10

pH = 10 -log 5.31 = 9.2749

d.

1.00E-2 mole of NH3 which is your base in the buffer
1.00E-2 mole of NH4Cl which is your acid in the buffer
Kb of NH3 is 1.8E-5

pH=pKa+log([base]/[acid])

to find the pka take the pKb which is the negative log of the Kb value and subtract it from 14 to get this

pKa=14-(-log(1.8E-5))
pKa=14.00-4.74
pKa=9.255

lets find the original pH in order to have something to compare our other answer to

pH=9.255+log(1.00E-2/1.00E-2)
pH=9.26