OpenStudy (anonymous):

You weigh a sample of potassium hydrogen phthalate (KHP) to prepare a primary standard for a titration. You later discover that the KHP was contaminated with sugar. To determine the amount of KHP in the mixture, you take 6.336 g of the mixture and make a 100.0 mL solution. You then titrate 10.00 mL of this solution with a 0.1787 M sodium hydroxide solution. You find that 13.75 mL of the NaOH solution is needed to reach the endpoint. What is the percent by mass of KHP in the mixture?

2 years ago
OpenStudy (anonymous):

we have Veq = 13.75 ml of (0.1787 M) we do : C1V1 = C2Veq ===> C1 = (C2Veq)/V1 AN : C1 = (0.1787 * 13.75)/ 10 = 0.2457 M so in 1000ml we have 0.2457 mol of KPH but we worked on 100ml so n = 0.02457mol === > m(KPH)= n*M(KPH) = 0.02457 * M finally 6.336 g ----> 100% and m(KPH) -----> X % ................done ........... sorry about my english writing .......

2 years ago
OpenStudy (anonymous):

@ljones69

2 years ago
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