You need to prepare 100 mL of a 0.5 M solution of potassium acetate (KCH2COOH).

ID: 1034937 • Letter: Y

Question

You need to prepare 100 mL of a 0.5 M solution of potassium acetate (KCH2COOH). You have 50 mL of a 0.75 M KCH2COOH solution and 30 mL of a 5% (w/v) KCH2COOH solution. What volume of 5% KCH2COOH and water needs to be added to the 50 mL of 0.75 M KCH2COOH solution to make up 100 mL of 0.5 M KCH2COOH? Molar mass of KCH2COOH = 98.15 g/mol % weight / volume, % (w/v) means mass (in g) in volume (100 mL)

Show your working for solving this question and complete the table below

Chemical Solution Volume (mL) 0.75M KCH2COOH 50.00 5% KCH2COOH H2O

Explanation / Answer

First, we will convert 5%(w/v) into molarity (M)

5%(w/v) KCH2COOH solution means in 100 mL of solution contains 5 g of KCH2COOH. Therefore, 1000 mL of solution has (5 g*1000 mL) /100 mL = 50 g of KCH2COOH

Molecular weight of KCH2COOH = 98.1423 g/mol

i.e., 98.1423 g of KCH2COOH = 1 mol of KCH2COOH

so, 50 g of KCH2COOH = 50/98.1423= 0.5095 mol

so, 0.5095 mol of KCH2COOH is present in 1000 mL of solution. Therefore, molarity is 0.5095 M

Thus, 5 % (w/v) KCH2COOH is equivalent to 0.5095 M of KCH2COOH -------(1)

Final concentration of KCH2COOH required = 0.5 M

Final volume of KCH2COOH required = 100 mL

mmol of KCH2COOH in final solution = 0.5 M * 100 mL = 50 mmol -----(2)

mmol of KCH2COOH given by 50 mL of 0.75 KCH2COOH = 0.75 M * 50 mL = 37.5 mmol ----(3)

therefore, mmol of KCH2COOH required from 5% KCH2COOH solution = 50 mmol - 37.5 mmol = 12.5 mmol

so, we need to find what volume of 0.5095 M or 5% (w/v) KCH2COOH will give 12.5 mmol of KCH2COOH

which is = mmol required / molarity of solution = 12.5 mmol / 0.5095 M = 24.53 mL

Thus, we need 24.53 mL of 0.5095 M or 5%(w/v) KCH2COOH solution

Amount of water required = Final volume - volume of 0.75 M solution - volume of 5%(w/v) solution

= 100 - 50 - 24.53 = 25.47 mL

Hence, Volume of water required = 25.47 mL

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You need to prepare 100 mL of a 0.5 M solution of potassium acetate (KCH2COOH).

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