Calculate the pH and the pOH of an aqueous solution that is 0.030 M in HCl(aq) and 0.095 M in HBr(aq) at 25 C
Thank you in advance
Total [H+] = [H+] from HCl + [H+] from HBr
= 0.030 M + 0.095 M
= 0.125 M
use:
pH = -log [H+]
= -log (0.125)
= 0.903
POH = 14 - pH
= 14 - 0.903
= 13.097
Answer:
pH = 0.903
POH = 13.097
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