Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH.

For 270.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH.

For 270.0 mL of a buffer solution that is 0.240 M in HCHO2 and 0.310 M in KCHO2, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH(Ka=1.8⋅10−4).

For 270.0 mL of a buffer solution that is 0.2604 M in CH3CH2NH2 and 0.2304 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0150 mol of NaOH(Kb=5.6⋅10−4).

Answer 1

1)

i) Initial pH of pure water will be 7.

ii)NaOH on addition to water dissociates as, NaOH(s) + H2O ------------- > Na+(aq.) + OH^-(aq.)

For aq.NaOH solution formed,

V=270.0 mL, [NaOH] = C = 0.0150 M, [OH^-] =?

NaOH is strong electrolyte which dissociates to the complete extent and hence,

[OH^-] = [NaOH] = 0.0150 M

Let us calculate the concentration in terms of M/L

0.0150 M of NaOH added to 270 mL of water, hence,

[OH^-] = 0.0150 M / 270 mL

So, [OH^-] = 5.56 x 10^-5 M/mL

So, [OH^-] = (5.56 x 10^-5) x 10³ M/L

[OH^-] = 5.56 x 10^-2 M/L ……………………… (Anathor way -From this we can find pOH and subtract it from 14 to get pH)

We know that,

[OH^-][H⁺] = Kw = 1 x 10^-14

[OH^-][H⁺] = 1 x 10^-14

(5.56 x 10^-2) [H⁺] = 1 x 10^-14

[H⁺] = (1 x 10^-14) / (5.56 x 10^-2)

[H⁺] = 1.8 x 10-13

Now, pH = -log [H⁺]

pH = -log(1.8x 10^-13)

pH = 12.75

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2)

i)Let initial pH be pH1.

For this solution [CHO2^‑] : [HCHO2] = 0.31: 0.24,

Ka = 1.8 x 10-4

So, pKa = -log(ka)

pKa = - log(1.8 x 10^-4)

pKa = 3.7447

According to Henderson’s equation,

pH = pKa + log{[conjugate base or salt]/[acid]}

Hence here we have,

pH1 = pKa + log{[CHO2^-]/[HCHO2]}

pH1 = 3.7447 + log(0.31/0.24)

pH1 = 3.7447 + 0.1112

pH1 = 3.8559

ii)Let pH after adding NaOH be pH2.

Addition of strong base like NaOH decrease the concentration of acid and increases the concentration of conjugate base (i.e. salt)

Let us find new concentrations,

Concentration of NaOH = 0.0150 x 270 = 4.05 mmol of NaOH

Originally, [HCHO2] = 0.240x270 = 64.8 mmol

After adding 4.05 mmol NaOH, [HCHO2] = 64.8 – 4.05 = 60.75

Originally, [KCHO2] = 0.310 x 270 = 83.7 mmol

After adding NaOH [CHO₂^-] = 83.7 + 4.05 = 87.75 mmol.

Hence new,   [CHO2^‑] : [HCHO2] = 87.75 : 60.75

By Henderson’s equation,

pH2 = 3.7447 + log(87.75/60.75)

pH2 = 3.7447 + 0.1597

pH2 = 3.9044

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3)

Given buffer is Ethyl amine (EA)/ethylammoniumchloride(EAcl) and it’s a basic buffer. We will calculate first pOH and then pH.

i) Let initial pH be pH1,

We have, Kb = 5.6 x 10-4

so, pKb = -log (Kb) = 3.2518

[EAcl]:[EA] = 0.2304:0.2604

Henderson’s equation for Basic buffer,

pOH = pKb + log{[conjugate acid]/[Base]}

pOH = pKb + log{[EAcl]/[EA]}

pOH1 = 3.2518 + log(0.2304/0.2604)

pOH1 = 3.2518 + (-0.0532)

pOH1 = 3.1986

hence,

pH1 = 14 – pOH1

pH1 = 14 – 3.1986

pH1 = 10.8014.

ii) After adding strong base NaOH, [EAcl] will decrease and that of [EA] will increase.

Let us find new concentrations, after adding 0.0150 M to 270 mL (i.e. 4.05 mmol) of buffer.

Originally, [EAcl] = 0.2304 x 270 = 62.21 mmol

After adding NaOH, [EAcl] = 62.21 – 4.05 = 58.16 mmol.

Originally [EA] = 0.2604 x 270 = 70.31 mmol.

After adding NaOH [EA] = 70.31 + 4.05 = 74.36 mmol

Hence, new [EAcl] : [EA] = 58.16:76.36

Henderson’s equation,

pOH2 = pKb + log(58.16/76.36)

pOH2 = 3.2518 + (-0.1182)

pOH2 = 3.1336

pH2 = 14 – 3.1336

pH2 = 14-3.6265

pH2 = 10.8664

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