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Calculate the [OH-] and pH of the solution?
Calculate [OH-] and pH of a 0.10 M NaCN solution. HCN (Ka = 4.9*10^(-10))
Do I write the equation as follows:
NaCN(aq) → Na+(aq) + CN-(aq)
Assuming that NaCN is an ionic compound (composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water, therefore [NaCN] = [CN-]
Then, since 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)) = 2.8
pH = 14 - pOH = 14 - 2.8 = 11.2
Is this solution correct? I tried doing it myself, but I'm pretty unsure of my work, so I would appreciate if someone could verify it. I have a test on Monday, but it would be nice if I could get immediate feedback on how I'm doing.
I suppose since it's a weak acid, I could also use the Henderson–Hasselbalch equation:
pH = pKa + log([conj.base] / [acid])
This could also be used with moles, I believe, but that wouldn't help me in this problem.
2 Answers
- 1 decade agoFavorite Answer
Well done, your working is correct.
A few tips though...you should always check that your assumption that x << 0.1 (i.e. x is no more than 1% of 0.1), which in this case it was (0.14%). Also, even though you are making the assumption, you should still present the difference in the equilibrium expression until you have made the statement you are going to make the assumption...so your 'E' concentrations should be:
(0.1 - x).. | . . . . x . . . . . x
and kb = x² / (0.1-x)
it is better to show all your working so should something go wrong, your marker can see your train of thought and give you marks regardless of any mistake you might make.
- vierraLv 44 years ago
88mL * 5M = 0.40 4 mol 7bcfbacd98cfa6ce29d0ad821b14f9Cl 32mL * 8M = 0.256 mol 7bcfbacd98cfa6ce29d0ad821b14f9NO3 because of the fact the two 7bcfbacd98cfa6ce29d0ad821b14f9Cl and 7bcfbacd98cfa6ce29d0ad821b14f9NO3 are stable acids (aka they dissociate thoroughly into H+ and Cl and NO3), 0.40 4 moles of 7bcfbacd98cfa6ce29d0ad821b14f9Cl at equilibrium could make 0.40 4 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9; and 0.256 moles 7bcfbacd98cfa6ce29d0ad821b14f9NO3 produces 0.256 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 additionally. so upload those and there are 0.696 moles. divide via the quantity, 1L, and it is 0.696 M 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 (first answer). the O7bcfbacd98cfa6ce29d0ad821b14f9- concentration is comparable to the quantity (10^-14)/[7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9]. so (10^-14)/(0.696) = a million.40 4 x 10^-14 M for the O7bcfbacd98cfa6ce29d0ad821b14f9- concentration. the p7bcfbacd98cfa6ce29d0ad821b14f9 is the neg-log of 0.696, it is comparable to a p7bcfbacd98cfa6ce29d0ad821b14f9 of 0.157 (it is vitally low, yet it is because of the fact it is stable acids and no bases to counter it)
