PLEASE PLEASE HELP WITH THESE! THANK YOU IN ADVANCE! You have a 1.153 g sample o

Question

PLEASE PLEASE HELP WITH THESE! THANK YOU IN ADVANCE!

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation HA(ag)-KOH(ag)- KA(aq) +H2O(1) 2nd attempt Part 1 (0.5 point) Feedback hi See Periodic Table See Hint If 13.40 mL of 0.710 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? 0.411 Part 2 (0.5 point) Feedback ? See Hint What is the molar mass of HA? 3 140.3 g/mol

Explanation / Answer

Question 1

Moles of KOH is used for titration = 13.4 x 0.71 / 1000 = 0.009514 Moles

Moles of HA present in 20 ml of sample = 0.009514 Moles

Concentration of the acid = 0.009514 x 1000 /20 = 0.4757 M

Molar mass of the acid = 1.153 gm / 0.00951 Moles = 121.24 g/mol

Question 2

Balanced equation:
MgO(s) + 2 HCl(aq) ====> MgCl2(s) + H2O(l)
Reaction type: double replacement

Moles of MgO = 0.71 / 40.30 = 0.017615 Moles

Moles of HCl needed =  0.03523 Moles

Volume of HCl needed = 0.03523 x 1000 /0.1 = 352.3 ml

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