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70 ml of hcl is titrated with a solution of m naoh it requires 101 ml of naoh to reach the equivalence point what is the concentration of the hcl solution?
Wiki User
∙ 13y ago
PH =- log (6.2*10^-10) + log (base/acid) Ka for HCN is (6.2*10^-10)
6.2*10^-10=X^2/0.1
X= 7.874*10^-6 M HCN
7.874*10^-6 *(57.4 ml/1000)=4.52*10^-7 M
(4.52*10^-7 M)/(7.874*10^-6M)=0.0574M concentration the base
0.1*(57.4/1000)=.00574 moles acid
now I have to find Kb
Kb = Kw/Ka
=1*10^-14/(6.2*10^-10) =1.613*10^-5
1.613*10^-5=X^2/.00574
X= 3.043*10^-4
-log( 3.043*10^-4)= 3.52
14 - 3.52=10.48 (my pH is wrong. It suppose to be 10.95) I don't know how I can
get the correct answer. Please can any body help me? Thanks alot.
Wiki User
∙ 16y ago
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