Multiple ChoiceCalculate the volume (mL) of 0.500 M NaOH required to reach the equivalence point in the titration of 25.0 mL of 0.650 M HF. Ka for HF = 3.5 × 10−4.548views
Multiple ChoiceFor a titration of a weak acid with a strong base, the pH at the equivalence point ________.439views
Multiple Choice20.0 mL of 0.500 M HC2H3O2 is titrated with 0.350 M KOH. Determine the pH after 15.0 mL of the base has been added. Ka for HC2H3O2 = 1.8 × 10−5.293views
Multiple Choice20.0 mL of 0.500 M HC2H3O2 is titrated with 0.350 M KOH. Determine the pH after 15.0 mL of the base has been added. Ka for HC2H3O2 = 1.8 × 10−5.263views
Multiple ChoiceDetermine the pH when 5.00 mL of excess KOH (beyond the equivalence point) has been added to the titration of 20.0 mL of 0.500 M HC2H3O2 with 0.350 M KOH.433views
Multiple ChoiceCalculate the pH at the equivalence point for the titration of 20.0 mL of 0.500 M HC2H3O2 with 0.350 M KOH. Ka for HC2H3O2 = 1.8 × 10−5.332views
Multiple ChoiceIn order to create a buffer 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? Ka = 4.9 × 10−10.293views4rank1comments
Multiple ChoiceConsider the titration of 75.0 mL of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The Ka of HNO2 is 4.6 × 10−4.293views3rank
Multiple ChoiceCalculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH. 251views4rank1comments
Open QuestionCalculate the pH when 50.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10-9)138views