Question

What volume (to the nearest 0.1 mL) of 6.90-M NaOH must be added to 0.500 L of 0.150-M HNO2 to prepare a pH = 3.80 buffer? 3.1 Incorrect: Your answer is incorrect. mL

Answer 1

Let the volume of NaOH added = x L

Moles of NaOH = molarity * volume

= 6.90 x

Moles of HNO2 = molarity * volume

= 0.150 mol/L * 0.500 L

= 0.075 mol

The balanced reaction with ICE TABLE

OH- + HNO2 ---> NO2- + H2O

I 6.90x 0.075

C - 6.9x - 6.9x +6.9x +6.9x

E 0 (0.075-6.9x) (6.9x) (6.9x)

From the Henderson-Hasselbalch equation

pH = pKa + log [base/acid]

3.80 = 3.40 + log [NO2-]/[HNO2]

3.80 = 3.40 + log [6.9x]/[0.075-6.9x]

0.40 = log [6.9x]/[0.075-6.9x]

2.512 = [6.9x]/[0.075-6.9x]

2.512[0.075-6.9x] = [6.9x]

0.1884 - 17.3328x = 6.9x

0.1884 = 24.2328x

x = 0.00777 L x 1000 mL/L

= 7.8 mL