Question

FIND THE PH AN OH- OF EACH. (Round to the second decimal place.)

a. 5.0×10⁻² M NaBrO.

b. 8.0×10⁻² M NaHS.

c. A mixture that is 0.13 M in NaNO2 and 0.22 M in Ca(NO2)2.

Answer 1

(a) Given that, [NaBrO] = 5 * 10^-2M = 0.05 M

Reaction is:

NaBrO + H₂O HBrO + NaOH

In ionic form

BrO^- + H₂O HBrO + OH^-

0.05 0 0

0.05-x x x

Thus, K_b = [OH^-][HBrO] / [BrO^-]

From literature, K_b = 4.26 * 10^-6

4.26 * 10^-6 = x.x / (0.05 -x) = x²/( 0.05 -x)

(2.13 * 10 ^-7) - (4.26 * 10^-6 )x = x²

x = 4.59 * 10^-4 = [OH^-]

pOH = -log [OH^-] = -log (4.59 * 10^-4) = 3.34

pH = 14 -pOH = 14 -3.34 = 10.66

(b) Given that, [NaHS] = 8 * 10^-2M = 0.08 M

Reaction is:

NaHS + H₂O HSH + NaOH

In ionic form

SH^- + H₂O HSH + OH^-

0.08 0 0

0.08-x x x

Thus, K_b = [OH^-][HBrO] / [BrO^-]

From literature, K_b = 1.122 * 10^-7

1.122 * 10^-7= x.x / (0.08 -x) = x²/( 0.08 -x)

(8.98 * 10 ^-9) - (1.122 * 10^-7)x = x²

x = 9.47* 10^-5 = [OH^-]

pOH = -log [OH^-] = -log ( 9.47* 10^-5 ) = 4.02

pH = 14 -pOH = 14 -4.02 = 9.98

(c) 0.13 M NaNO₂ = 0.13 M NO₂^-

and 0.22 M Ca(NO₂)₂ = 0.44 M NO₂^-

Thus, total [NO₂^-] in mixture = 0.44 + 0.13 = 0.57 M

Reaction is:

NO₂^- + H₂O HNO₂ + OH^-

0.57 0 0

0.57 -x x x

From literature , K_b = 2.5 * 10^-11

K_b = [HNO₂^-] [ OH^-] / [NO₂^- ]

2.5 * 10^-11 = x²/ (0.57 -x)

( 1.425 * 10^-11 ) - (2.5 * 10^-11 )x = x²

x = 3.77 * 10^-6 = [OH^-]

pOH = -log [OH^-] = -log ( 3.77 * 10^-6) = 5.42

pH = 14 -pOH = 14 -5.42 = 8.58