Answer:
c. 100:1.
Explanation:
Hello,
In this case, considering the Henderson-Hasselbach equation:
[tex]pH=pKa+log(\frac{[A^-]}{[HA]} )[/tex]
And the pH two units above the pKa:
[tex]pH=pKa+2[/tex]
We obtain the required ratio as follows:
[tex]pKa+2-pKa=log(\frac{[A^-]}{[HA]} )\\\\\frac{[A^-]}{[HA]} =10^2=100[/tex]
Which is also a c. 100:1 ratio containing 100 parts of base per 1 part of acid.
Regards.
According to Henderson - Hawelbach equation:
[tex]\to pH = pka + \log (\frac{A^{-}}{HA})...........(a)[/tex]
According to question pH two unit above pka:
[tex]\to pH = pka + 2[/tex]
Put value of PH is equation (a)
[tex]\to pKa+2=pKa+ \log (\frac{A^{-}}{HA})\\\\\to 2= \log (\frac{A^{-}}{HA})\\\\\to (\frac{A^{-}}{HA})= 10^2\\\\\to (\frac{A^{-}}{HA})= 100\\\\[/tex]
So, the ratio is [tex][100 : 1][/tex]
Therefore, 100: 1 Ratio containing 100 part of base per 1 part of acid.
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