Question

Given 2.5 M HNO₂ , 5.0 M NaOH, and DI water, what are the volumes necessary to create 500 mL of 1.56 M, pH 4.5, nitrous acid buffer. (Ka of HNO₂ = 4.0 x 10^-4). [step by step solution please]

Answer 1

HNO2 + NaOH -----> NaNO2 + H2O

pH of he buffer is given as

pH = pKa + log[sal]/[acid]

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

pKa = -logKa = - log 4.0 x 10^-4= 3.397,

pH = 4.5

4.5 = 3.397 +log[NaNO2]/[HNO2]

log[NaNO2]/[HNO2] = 4.5 - 3.397 = 1.103

or [NaNO2]/[HNO2] = - antilog (1.103) = 0.07888

[NaNO2] = [HNO2] 0.07888

[NaNO2] + [HNO2] = 1.56 ...(Given)

[HNO2] 0.07888 + [acid] = 1.56

1.07888[HNO2] = 1.56

[HNO2] = 1.56 / 1.0788 = 1.446M

therefore[NaNO2] = 1.56 - 1.446 = 0.114M

Thus the concentration of HNO2 must be 1.446M and NaOH concentration must be 0.114M

Because the amount of NaNO2 is equal to the amount of NaOH added.

Volumes needed to make desired concentration can be calculated as

M1V1 = M2 V2

M1V1 = molarity and volume of stock solution and M2V2 = Moarity and concentration of solution to be prepared

(a) HNO2 solution

2.5 X V1 = 1.446 X 500

V1 = 289.2mL

Hence 289.2mL of stock solution having molarity 2.5 must be mixed with 500-289.2 = 210.8mL of distilled water to make 500mL of 1.446M HNO2

(b) NaOH

5.0 X V1 = 0.114 X 500

V1 = 11.4mL

Thus 11.4mL of 5.0M NaOH solutiom must be mixed with 500-11.4 = 288.6mL of distilled waterto make 500mL of 0.114M NaOH solution