A) 4 : 1
B) 7 : 1
C) 5 : 1
D) 2 : 1
Correct Answer: C
Solution :
At half-way between two colours \[[H\,\,In]=[I{{n}^{-}}]\] |
Now \[pKIn=pH+\log \frac{[H\,\,In]}{[I{{n}^{-}}]}=pH\] |
Now pH of buffer = 5.45 when |
\[5.45=4.75+\log \frac{[\text{salt }\!\!]\!\!\text{ }}{[\text{acid }\!\!]\!\!\text{ }}\] |
So \[\log \frac{[\text{salt }\!\!]\!\!\text{ }}{[\text{acid }\!\!]\!\!\text{ }}=0.70\] |
\[\frac{[\text{salt }\!\!]\!\!\text{ }}{[\text{acid }\!\!]\!\!\text{ }}=5:1\] |
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