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If the pKa1 value for H2CO3 is 6.4, what is the pH of
a 103 M solution of this acid?
A . 3.0
B . 4.7 ---correct ans
C . 6.3
D . 9.3
Ka will be 10^-6.4 and thus Ka = (HCO3-)(H+)/ (H2CO3)
10^-6.4 = x2/ 10^-3
x2 = 10^9.4
x approximately equal to 3, infact a little less than 3.
Can anyone please solve this problem. I don't understand the Kaplan explanation. They said that the ka = 4*10^-7
a 103 M solution of this acid?
A . 3.0
B . 4.7 ---correct ans
C . 6.3
D . 9.3
Ka will be 10^-6.4 and thus Ka = (HCO3-)(H+)/ (H2CO3)
10^-6.4 = x2/ 10^-3
x2 = 10^9.4
x approximately equal to 3, infact a little less than 3.
Can anyone please solve this problem. I don't understand the Kaplan explanation. They said that the ka = 4*10^-7