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55 mL of 0.0380 M KOH and 45 mL of 0.0530 M HNO3 are mixed together....

55 mL of 0.0380 M KOH and 45 mL of 0.0530 M HNO3 are mixed together. Find the pH of the mixture after they react.
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Answer #1

millimoles of KOH = 55 x 0.0380 = 2.09

millimoles of HNO3 = 45 x 0.0530 = 2.385

millimoles of acid > millimoles of base

millimoles of acid remains = 2.385 - 2.09 = 0.295

acid concentration remains = 0.295 / total volume

                                            = 0.295 / (55 + 45)

                                            = 2.95 x 10^-3 M

[H+] = 2.95 x 10^-3 M

pH = -log [H+]

pH = -log (2.95 x 10^-3 )

pH = 2.53

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