Titration of a Weak Acid with Strong Base?

Question

70.0 mL of 0.100-M hydrogen cyanide (Ka = 6.20×10-10) is titrated with 0.100-M NaOH.

What is the initial pH of the hydrogen cyanide solution?


I have such a problem with titrating problems...Help please..

Answer 1

HCN is a weak acid. U can tell from the Ka value that is much <<1.

equation :HCN <-> H+ + CN-
initial con:0.1M....... 0....... 0
change... : -x ..........+x..... +x
equilibrm : 0.1-x.......x.........x

Ka=6.2x10-10 = x.x/ 0.1-x

By solving the value of x, u get the concentration of H+. Using -log[H+], u get the INITIAL pH

Answer 2

Since [H+] for HCN = root (Ka x 0.100), you can easily work out the initial pH of the HCN.

The above answerer has made it an alkali...

Answer 3

I have answered but I can answer again
Initial solution Has only HCN slighty dissociated

HCN <> H+ + CN-
initial concentration
0.100......0........0
at equilibrium
0.100-x....x........x

6.20 10^-10 = (x)(x) / 0.1 -x

x = 0.00000787 M

pH = -log 0.00000787 = 5.10

After adding e.g. 10 mL of NaOH 0.100M
70 mL HCN gives 0.07 L(0.100 mole) = 0.007 mole HCN

10 mL of NaOH 0.100 M gives 0.00100 mole OH-

Total volume = 80 mL=0.08 L

0.07 mole HCN + 0.00100 mole OH- >> 0.006 mole HCN + 0.00100 mole CN-

Concentration HCN = 0.006 /0.007 = 0.857M
Concentration CN_ = 0.00100 / 0.007 = 0.143 M

HCN<> H+ + CN-
at equilibrium
0.857-x...x........x+0.143

6.20 10^-10 = x(x+0.143) / 0.857-x

x= 4 10^-9

pH= 8.43