Question

A student dilutes 50 mL of a phosphoric acid-dihydrogen phosphate buffer solution to a final volume of 100 ml.

Will the pH of the buffer change noticeably? (Use Henderson -Hasselbeck conceptually )

For the same dilution above will the buffer capacity change ?

Calculations: A 50.0 mL sample of a 0.100 M phosphoric acid, H3PO4 is titrated with 0.100 M NaOH.

Identify the major chemical species present and calculate the pH at these four points (first consider the titration curve)

Given : K_a (H₃PO₄)=7.5 x10^-3, K_a(H₂PO₄^-)=6.2x10^-8, K_a(HPO₄^2-)=4.2x10^-13

^{Point 1: after 20.0 mL of NaOH is added}

^{Point 2: after 50.0 mL total of NaOH is added}

^{Point 3: after 75.0 mL total of NaOH is added}

^{Point 4: after 100.0 mL total of NaOH is added}

Answer 1

Will the pH of the buffer change noticeably? (Use Henderson -Hasselbeck conceptually )

No, since according to the equation:

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

since this is a dilution to the same volume, the concentration ratio is done... i.e. Volume cancels each other.

For the same dilution above will the buffer capacity change ?

Yes... buffer capacity slightly, since it is much dilute

mmol of acid = MV = 50*0.1 = 5 mmol of acid --> 5*3 = 15 mmol of H+

pKa1 = -log(Ka1) = -log(7.5*10^-3) = 2.1249

pKa2 = -log(Ka1) = -log(6.2*10^-8) = 7.21

pKa3 = -log(Ka3) = -log(4.2*10^-13) = 12.376

Point 1:

mmol of NaOH = MV = 20*0.1 = 2 mmol of base

then..

mmol of H2PO4 fomred = 2

mmol of H3PO4 left = 5-2 = 3

pH = pKa1 + log(2/3) = 2.1249 + log(2/3) = 1.95

Point 2:

mmol fo NaOH --> 50*0.1 = 5 mmol

first equivalence point is reached:

5 mmol of H3PO4 + OH- --> H2O + H2PO4-

pH2 = (pKa1+pKa2)/2 = (2.1249+7.21)/2 = 4.667

point 3:

mmol of OH_ = MV = 75*0.1 = 7.5 mmol

5 mmol used for H3PO4, 2.5 mmol of OH- left

5-2.5 = 2.5 mmol of H2PO4- left

pH = pKa2 + log(conjguate-7acid)

pH = 7.21 + log(2.5/2.5) = 7.21

point 4:

mmol of OH- = MV = 100*0.1 = 10 mmol

this is the 2nd ionization point

pH = (pKA2+pKa3)/2 = (7.21+12.376)/2 = 9.793