Hi Expert! Please answer all the following parts: Answer all parts: A) Potassium
ID: 902754 • Letter: H
Question
Hi Expert!
Please answer all the following parts:
Answer all parts:
A) Potassium Hydrogen Phthalate (KHP), C6H4C2O4KH (FW 204.221), is a primary standard used to standardize sodium hydroxide. Calculate the molarity of a sodium hydroxide solution that required 27.51 mL to react with 0.4982 g of KHP.
B) The water hardness (dissolved Ca2+ and Mg2+) of a 50.00 mL sample of GSU tap water (era 2003 before campus-wide water softening) was determined by direct titration with 27.53 mL of 0.0681 M EDTA. Calculate the water hardness, (molarity of sum of dissolved Ca2+ and Mg2+), for the GSU tap water sample.
C) A 50.00 mL sample of a solution containing Co2+ was treated with 25.00 mL of 0.0542 M EDTA to complex all the Co2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 8.48 mL of 0.035 M Zn2+. What was the concentration of Co2+ in the original solution?
Thank you in advance!
+NaOHO + H,OExplanation / Answer
A) Potassium Hydrogen Phthalate (KHP), C6H4C2O4KH (FW 204.221), is a primary standard used to standardize sodium hydroxide. Calculate the molarity of a sodium hydroxide solution that required 27.51 mL to react with 0.4982 g of KHP.
Solution :-
Lets first calculate the moles of the KHP
Moles = mass/ molar mass
Moles of KHP = 0.4982 g / 204.221 g per mol = 0.00244 mol KHP
Mole ratio of the KHP to NaOH is 1 :1 therefore moles of NaOH reacted are same as moles of KHP
So moles of NaOH = 0.00244 mol
Now lets calculate the molarity of the NaOH
Molarity = moles/ volume in liter
Molarity of NaOH = 0.00244 mol / 0.02751 L
= 0.0887 M
So the molarity of the NaOH = 0.0887 M
B) The water hardness (dissolved Ca2+ and Mg2+) of a 50.00 mL sample of GSU tap water (era 2003 before campus-wide water softening) was determined by direct titration with 27.53 mL of 0.0681 M EDTA. Calculate the water hardness, (molarity of sum of dissolved Ca2+ and Mg2+), for the GSU tap water sample.
Solution :-
Lets first calculate the moles of the EDTA
Moles of EDTA = molarity * volume in liter
= 0.0681 mol per L * 0.02753 L
= 0.001875 mol EDTA
Moles of Ca^2+ and Mg^2+ total reacted with EDTA are same as moles of EDTA
So moles of cations Ca^2+ and Mg^2+ total = 0.001875 mol
Now lets calculate its molarity
Molarity = moles / volume in liter
= 0.001875 mol / 0.050 L
= 0.0375 M
So the hardness of water = 0.0375 M
C) A 50.00 mL sample of a solution containing Co2+ was treated with 25.00 mL of 0.0542 M EDTA to complex all the Co2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 8.48 mL of 0.035 M Zn2+. What was the concentration of Co2+ in the original solution?
Solution :- Lets first calculate the total moles of EDTA
Moles of EDTA = 0.0542 mol per L * 0.025 L = 0.001355 mol
Now lets calculate moles of Zn^2+
Moles of Zn^2+ = 0.035 mol per L * 0.00848 L = 0.000297 mol Zn^2+
So the moles of Co^2+ reacted = total moles of EDTA – moles of Zn^2+
= 0.001355 mol – 0.000297 mol
= 0.001058 mol
Now lets calculate the molarity of the Co^2+
Molarity of Co^2+ = 0.001058 mol / 0.05 L
= 0.0212 M Co^2+
So the concentration of the Co^2+ in the original solution is 0.0212 M