Problem 2 A 25.00 mL aliquot of HCI was titrated to the neutralization point at

Question

Problem 2 A 25.00 mL aliquot of HCI was titrated to the neutralization point at 20.20 mL of o 975 M NaoH what was the molarity of the HCI Answer M HCI Problem 3 A 25.00 mu aliquot of Hrso. was titrated to the neutralization point at 40.40 mL of 0.975 M NaOH. What was the molarity of the H2Soe? Answer M H2SOe Problem 4 A 25.00 mL aliquot of H PO. was titrated to the neutralization point at 60.60 mL of 0.975 M NaoH What was the molarity of the HoPOe? Answer M HPO. Problem 5 A 50.00 mL aliquot of Unknown Acid was titrat to the neutralization point at 27.41 mL of 0.975 M NaOH. What was the number of moles of the unknown Acid Answer moles Problem 6 A 8.4 grams of Unknown Acid [mono-protic was titrated to the neutralization point at 102.56 mL of 0.975 M NaOH [Hint: Molecular weight has units of grams/mole. 1 What was the number of moles of the Unknown Acid Answer moles what was the Molecular weight of the Unknown Acid Answer grams/mole

Explanation / Answer

Problem 2 -- HCl + NaOH ----> NaCl + H2O

From the above reaction it is clear that one mole of HCl require one mole of NaOH for complete neutralisation.

So we can conclude that for complete neutralisation --- moles of NaOH = moles of HCl

Moles = Molarity*Volume

Molarity of NaOH =0.975M

Volume of NaOH = 20.20mL

Moles of NaOH =0.975*20.20

Let Molarity of HCl be x

Volume of HCl =25.00mL

Moles of HCl =x*25.0

For neutralisation - Moles of NaOH =Moles of HCl

0.975*20.20 =x*25.0

x = 0.7878

Molarit of HCl is 0.7878M

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