what is the ph of a 0.10m solution of nacn at 25°c (ka=4.9×10^-10 for hcn). I've...

what is the ph of a 0.10m solution of nacn at 25°c (ka=4.9×10^-10 for hcn). I've gotten 5.15 as a pH but the correct answer is 11.15. I don't see how it's 11.15...

Expert Solution

HCN (Prussic Acid) is a weak acid, write the equilibrium equation between the acid and its conjugate base:
..... H2O(l) + CN-(aq) → HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored.

I ........ .......... 0.1 . . | . . . . 0 . . . . . 0
C ...... ............ -x . . | . . . .+x . . . . +x
E ...... .......... 0.1 . . | . . . . x . . . . . x

Since Ka was given, change it to Kb by using:
Ka * Kb = Kw
4.9*10^(-10) * Kb = 10^(-14)
Kb = 2.0*10^(-5)

Then set up the base dissociation constant equation (since it was written with a CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller than [CN-], just to make the calculations easier (avoiding quadratics).
Kb = [HCN] [OH-] / [CN-]
2.0*10^(-5) = x² / 0.1
x² = 2.0*10^(-6)
x = [OH-] = 1.4*10^(-3) M

pOH = -log[OH-] = -log(1.4*10^(-3)) = 2.85
pH = 14 - pOH = 14 - 2.85 = 11.15

queen_honey_blossom answered 1 year ago

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