Calculate the ph of a 0.170 m nh3-0.187 m nh4cl buffer when a.20.0 ml of 0.100 m hcl, b.20.0 ml of 0.100 m naoh, are added to 200 ml of buffer. ka nh4 + = 5.69 x 10-10

(a) 9.11; (b) 9.26

pH of original buffer

Step 1. Write the chemical equation for the equilibrium

Acid + H_2O ⇌ base + H_3O^(+) ; K_a = 5.69 × 10^(-10)  

Step 2. Calculate the pH of the buffer

The Henderson-Hasselbalch equation is pH = pK_a + log([base]/[acid]).

pH = -log[5.69 × 10^(-10)] + log[(0.170 mol)/L)/(0.187 mol/L)] = 9.24 + log0.1091

= 9.24 – 0.0414 = 9.20

(a) pH after adding HCl

Step 1. Calculate the moles of acid and base in the buffer

Moles of acid = 200 mL × (0.187 mmol/1 mL) = 37.4 mmol

Moles of base = 200 mL × (0.170 mmol/1 mL) = 34.0 mmol

Step 2. Calculate the moles of HCl  added

Moles of HCl added = 20.0 mL × (0.100 mmol/1 mL) = 2.00 mmol

Step 3. Calculate the new moles of acid and base

New moles of acid = (37.4 + 2.00) mmol = 39.4 mmol

New moles of base = (34.0 – 2.00) mmol = 32.0 mmol

Step 4. Calculate the new pH

pH = 9.20 + log[(32.0 mmol)/(39.4 mmol)]) = 9.20 + log0.8122 = 9.20 - 0.0903   = 9.11

(b) pH after adding NaOH

Step 1. Calculate the moles of NaOH added

Moles of NaOH added = 20.0 mL × (0.100 mmol/1 mL) = 2.00 mmol

Step 2. Calculate the new moles of acid and base

Moles of acid = (0.187 - 0.0200) = 0.167 mol

Moles of base = (0.170 + 0.0200) = 0.190 mol

Step 3. Calculate the new pH

pH = 9.20 + log[(0.190 mol/L)/(0.167 mol/L)]) = 9.20 + log1.138 = 9.20 + 0.0560   = 9.26

i think   the answer is c hope i'm right