What is the pH if 50.0 mL of 0.100 M HCN is mixed with 100.0 mL of 0.100 M NaOH?

Given: K_a for HCN is 1.00xx10^-6

I'm asked to work with that value

Attempting first ICE table:
HCN+NaOH\toH_2O+NaCN
(I) 0.0050 " " 0.0100 " " " || " " " 0
(C)-0.0100 " -0.0100 " " || " " " +0.0100
(E)-0.0050 " " " " 0 " " " || " " " 0.0100

1 Answer
Jul 5, 2018

"pH" = 12.523

Explanation:

K_b -= K_w / K_a = 1.00 xx 10^(-8)

The strong base "NaOH" is in excess and neutralizes all "HA" initially present in the solution. As a result, the hydrolysis of "A"^(-), the product of the neutralization reaction. would be the only reversible process capable of generating an equilibrium.

"A"^(-)(aq) + "H"_2"O"(aq)\ color(purple)(rightleftharpoons) \ "HA"(aq) + ul("OH"^(-) (aq))

R "A"^(-)(aq) + "H"_2"O"(aq)\ color(purple)(rightleftharpoons) \ "HA"(aq) + ul("OH"^(-) (aq))
I 0.005 color(white)(------------) 0.005
C -x color(white)(-------ll)+x color(white)(----) +x
E 0.005 -x color(white)(-----l-)x color(white)(---ll) 0.005 +x

(in mol)

c = n /color(purple)(V) and therefore

K_b("A"^(-)) = (overbrace(["HA"])^("conjugate acid") * ["OH"^(-)])/(overbrace(["A"^(-)])^("weak base"))\ = ((x)/(0.150)*(0.005+x))/(0.005 -x )

(x*(0.005+x))/(0.005 -x ) = 1.00 xx 10 ^(-8) * 0.150

x^2 + (0.005 + 1.00 xx 10^(-8))* x -0.005 * 1.00 xx 10 ^(-8) = 0

x = 1.50 xx 10 ^(-9) color(white)(l) mol

(x > 0 given that ["HA"] >= 0)

Therefore

n("OH"^(-)) = 0.005 + 1.50 * 10^(-9) ~~ 0.005 color(white)(l) mol
c("OH"^(-)) = n / V = 0.0333 color(white)(l) mol * dm^(-3)

"pH" = "pKw" - "pOH" = 12.523

Note that "OH"^(-), already present at great quantity, is the only factor through which the equilibrium influences the "pH" of the solution. K_b = 1.00 xx 10^(-8); thus there's a small deviation in "pH" from what would have been observed with a strong acid of the same volume and concentration.