Suppose you titrate 0.200 L of a 0.150 M solution of NaNO2 with 6.0 M HCl. Ka for HNO2 is 5.6 x 10–4 Need b/c?
(a) What is the pH of the solution before beginning the titration?
(b) What is the pH of the solution half-way through the titration?
(c) What is the pH at the equivalence point?
a. NaNO2 + H2O = HNaNO2+OH-
.150 - 0 0
-x +x +x
.150-x x x
pkb = 1.79x10-11= x^2/.150
√(x^2 = 2.68x10-12)
x = OH- = 1.64x10^-6
pOH = -log (1.65x10^-6) 5.79
pH = 14-5.79 = 8.21
b and c is where I get lost
I hope I did A right
But I really need to figure out how to do the last two parts if someone can help.
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Verified answer
a) is correct
pOH = 0,5*(pKb - lg(0,15)) = 5,8
=>
pH = 8,21
===
b) at half-way through the titration, we have n(NO2^-) = n(HNO2) =>
pH = pKa = 3,25
===
c)
at equivalence point we have: n(A^-) = n(H^+). But we need V(HCl) to calculate pH exact.
we have established 0,15 M HNO2 at eq.
pH = 0,5*(pKa - lg(0,15)) = 2,04