Question

What volume (to the nearest 0.1 mL) of 6.00-M NaOH must be added to 0.400 L of 0.250-M HNO₂ to prepare a pH = 4.30 buffer?

What volume (to the nearest 0.1 mL) of 5.50-M HCl must be added to 0.250 L of 0.200-M K₂HPO₄ to prepare a pH = 7.10 buffer?

Answer 1

The reaction between NaOH and HNO₂ can be written as

Moles of HNO2 is 0.4 x 0.25 = 0.1

NaOH + HNO₂ NaNO2 + H₂O ICE table
initial: (6.00 x x1mL) (0.1) 0 0 (mol)
delta: (-6.00x1) (-6.00x1) (6.00x1) (6.00x1) (mol)
eq: (0.0) (0.1 - 6.00x1) (6.00x1) (6.00x1) (mol)


We know the equation:

pH = pKa + log(base/acid)
the pKa of HNO₂ is 3.40

so we get:

4.30 = 3.40 + log([NaNO₂]/[HNO₂])
4.30 = 3.40 + log((6.00x1)/(0.1 - 6x1))

log((6.00x1)/(0.1 - 6x1)) =0.9

(6.00x1)/(0.1 - 6x1) = 7.94

6x1 = 0.794 - 47.66x1

53.66x1 = 0.794

x1 = 0.794/53.66

x1 = 0.01479 L

14.8 mL of 6.0 M NaOH is required

The reaction between HCl and K₂HPO₄ is

moles of K₂HPO₄ is 0.25 x 0.2 = 0.05

HCl + K₂HPO₄ KCl + KH₂PO4 ICE table

initial 5.5x1 0.05 0 0

change -5.5x1 -5.5x1 5.5x1 5.5x1

equilibrium 0 0.05-5.5x1 5.5x1 5.5x1

We know the equation:

pH = pKa + log(base/acid)
the pKa of K₂HPO₄  is 7.21

so we get:

7.1 = 7.21 + log([K₂HPO₄]/[KH₂PO₄])
7.1 = 7.21 + log([0.05-5.5x1]/[5.5x1])

log([0.05-5.5x1]/[5.5x1]) = -0.11

[0.05-5.5x1]/[5.5x1] = 0.776

0.05-5.5x1 = 4.27x1

0.05 = 9.77x1

x1= 0.05/9.77

x1 = 0.00511L

5.10 mL of 5.50-M HCl