Question #138102
Determine pH of a 4*10-3M Na-salicylate solution in 0.05 M NaCl .
1
Expert's answer
2020-10-15T03:21:09-0400

PH=7.20+lg(C1C2)=7.20+lg(0.0040.05)=7.20+lg0.08=7.20−1.09691001=6.10PH=7.20+lg(\frac{C1}{C2})=7.20+lg(\frac{0.004}{0.05})=7.20+lg0.08=7.20-1.09691001=6.10


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