Question

Calculate the pH of the solution that results from adding 20.0 mL of 0.200M NaOH to 50.0 mL of 0.135 M HNO2.

(The answer is 3.51)

Answer 1

HNO2 + NaOH ---> NaNO2 + H2O

Moles of HNO2 =Molarity*V in L = 50*0.135/1000 = 6.75 millimoles

Moles of NaOH = 0.2*20/1000 = 4 millimoles

HNO2 + NaOH ---> NaNO2 + H2O

6.75 4 0

6.75-4 0 4

Weak acid and weak salt are formed ----> which is a buffer system

Buffer formula

pH = pKa + log [salt]/[Base]

pKa of HNO2 = -log Ka = - log 4.5 x 10-4 = 3.34

pH = 3.34 + log [ 4 / 6.75-4 ]

pH = 3.502

U havent given Ka value if it have been given ...i woulg have got accurate answer

Thanking you