Question

g 30.0 mL of a buffer is made by mixing 20.0 mL of 0.10 M HNO2 and 10.0 mL of 0.10 M NaOH solution. What is the pH of this buffer after 0.04 mL of 10.0 M hydrochloric acid are added to it

Answer 1

pH = 2.98

Step-by-step explanation:

pKa of HNO2 is 3.35

HNO2 reacts with NaOH as follows:

HNO2 + NaOH → NaNO2 + H2O

Where moles of NaOH added are equal to NaNO2 produced

And initial moles of HNO2 - Moles NaOH = Moles HNO2

Initial moles HNO2:

0.0200L * (0.10mol/L) = 0.00200moles HNO2

Moles NaOH = Moles NaNO2:

0.0100L * (0.10mol/L) = 0.00100 moles NaOH = Moles NaNO2

Moles HNO2:

0.00200moles - 0.00100moles = 0.00100 moles HNO2

Then, the added HCl reacts with NaNO2 producing HNO2:

HCl + NaNO2 → HNO2 + NaCl

Where moles added of HCl are moles of HNO2 produced and initial moles of NaNO2 - moles of HCl = Moles NaNO2

Moles HCl:

4x10⁻⁵L * (10.0mol / L) = 0.00040moles HCl = Moles of HNO2 produced

Moles HNO2:

0.00100moles + 0.0004moles = 0.0014moles HNO2

Moles NaNO2:

0.00100moles - 0.0004moles = 0.0006moles NaNO2

With these moles and H-H equation we can solve for the pH of the buffer:

pH = pKa + log [NaNO2] / [HNO2]

Where pH is pH of the buffer

pKa is pKa of HNO2: 3.35

[] are moles of each species:

Replacing:

pH = 3.35 + log [0.0006moles NaNO2] / [0.0014moles HNO2]

pH = 2.98