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Sagot :
After 0.0015 mol of NAOH is added to 0.5000 l of this solution, the pH is 3.37.
Steps
First, modify the Henderson-Hassle-batch equation to find the acid's pka.
[tex]pH=pka + log\frac{[A]}{[HA]}[/tex]
[tex]pka = pH-log\frac{[A]}{[HA]}[/tex]
[tex]=3.35- log \frac{[0.1500]}{[ 0.2000 ]}[/tex]
[tex]pka =3.47[/tex]
Molarity of the added NaOH.
[tex]M=\frac{mol}{L}[/tex]
[tex]=\frac{0.0015 mol}{ 0.5000L}[/tex]
= 0.003 M
In water, NaOH will dissociate
[tex]NaOH- > Na^{+} + OH^{-}[/tex]
strong base OH will react with HA
HA+ OH --> [tex]H_{2} O[/tex] + A
the reaction also produced A. as, all OH are consumed in the reaction
HA will be decreased by 0.003 M and A will be increased by 0.003 M.
[tex]HA= 0.2000-0.003 \\= 0.1970[/tex]
[tex]A=0.1500+0.003=0.1530 M[/tex]
solving new pH using pka and the new values of [HA] and [A]
[tex]pH=pka + log\frac{[A]}{[HA]}[/tex]
[tex]=3.35 + log\frac{[0.1530]}{[0.1970]}[/tex]
pH=3.37 is the ph after 0.0015 mol of naoh is added to 0.5000 l of this solution
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