Please explain the mechanism to me as well.
[HOBr] = mol of HOBr / volume
= 0.50 mol / 1.00 L
= 0.50 M
[KOBr] = mol of KOBr / volume
= 0.30 mol / 1.00 L
= 0.30 M
Ka = 2*10^-9
pKa = - log (Ka)
= - log(2*10^-9)
= 8.699
use:
pH = pKa + log {[conjugate base]/[acid]}
= 8.699+ log {0.3/0.5}
= 8.477
Answer: 8.48
Please explain the mechanism to me as well. Part A Calculate the pH of the solution...
Calculate the pH of the solution made by adding 0.50 mol of HObr in 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HObr is 2.0 * 10^-9
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