Question

what is thE ph for

A. 0.021 M hno2 at 25 degrees with ka of 4.3*10^-4

B. .15 M nh2ch5 at 25 with kb of 5.4*10^-5

C. .45 m k2so4 at 25 with kb of 8.33*10^-13

Answer 1

what is thE ph for

A. 0.021 M hno2 at 25 degrees with ka of 4.3*10^-4

The pH of given acid can be calculated as

0.021 M HNO2 at 25 degrees The Ka of acid is = 4.3X10^-4

pH = -log[H+]

so let us find the concentration of H+

HNO2 <---> H+ + NO2-

HNO2 will dissociate x amount into x amount of H+ and OH-.

ka = 4.3 X 10^-4

ka = [H+][OH-]/[HNO2]

4.3X10^-4= x^2 / (0.21-x)

0.0000903 - 0.00043 x = x^2

x^2 - 0.0000903 + 0.00043 x = 0

On soving for x

x = [H+] = [NO2-] = 0.0092 M

so pH =-log [0.0092]

pH = 2.03

B. .15 M nh2ch5 at 25 with kb of 5.4*10^-5

pOH = -log[OH-]

CH3NH2 + H2O -> [OH-] + [CH₃NH₃⁺]

so let us find the concentration of OH-

CH3NH2 will hydrolysed x amount into x amount of CH3NH3+ and OH-.

kb = 5.4X 10^-5

kb = [CH3NH3+][OH-]/[CH3NH2]

5.4X10^-5= x^2 / (0.15-x)

0.0000081 - 0.000054 x = x^2

x^2 - 0.0000081 + 0.000054 x = 0

On soving for x

x = [OH-]= 0.00281 M

so pOH =-log [0.00281]

pOH = 2.55

pH= 14-2.55 = 11.45