Calculate the OH− concentration after 52.6
mL of the 0.100 M KOH has been added to
25.0 mL of 0.200 M HBr. Assume additive
Answer in units of M.
NO SIG FIGS, NEED FULL ANSWER
KOH + HBr =====> H2O + KBr
52.6 ml * 1 L / 1000 mL * 0.100 mol/L = 0.00526 mol KOH
25.0 ml * 1 L / 1000 mL * 0.200 mol/L = 0.005 mol HBr
KOH is in excess
After reaction occurs there are 0.00026 mol KOH in 77.6 mL of solution
[OH-] = 0.00026 / 77.6 mL * 1000 mL / 1 L = 0.00335 M
pOH = - log(0.00335) = 2.475
pH = 14 - 2.475 = 11.5251116267
Please best answer.
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