Ka = 7.5*10^-3
pKa = - log (Ka)
= - log(7.5*10^-3)
= 2.125
use:
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[H2PO4-]/[H3PO4]}
= 2.125+ log {0.89/0.50}
= 2.375
Answer: 2.38
12 Question 12 Tries remaining1 Points out of 1.00 lgquestion What is the pH of a solution that contains 0.50M H3PO4...
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