1approximately 2E-7 ??
This is very simple and basic question
The pH of aqueous solution of NaOH is given 7.3 at 298 K
Find the concentration on Na+ in the solution
Kw=10-14
Provide your method also
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8 Answers
1pOH= 14- 7.3 = 6.7
[OH-] = antilog(-6.7) = 2E-7
since its a strong base, there will be full dissociation.. so,
[Na+] = [OH-] = 2E-7 M
9i think if we consider OH due to water
H = 10^-7.3
OH = 10^-6.7 = H + Na
Na = 10^-6.7 - 10^-7.3 = 1.494 X 10^-7
1Since pH+pOH=ln[10^-14}=14
therefore the pH has to be the given value i.e 7.3
m i rite>>>>>>?????????pls scrap back
1357The answer given by Celestine is correct.
Here pH=7.3 so [H+]=5×10-8
[OH-]=Kw/[H+]=2×10-7
By conservation of charge
[H+]+[Na+]=[OH-]
[Na+]=1.5×10-7