Question

# How could you make 500 mL of a 0.1 M glycine HCl solution? the MW of...

How could you make 500 mL of a 0.1 M glycine HCl solution? the MW of glycine is 75.07 and assume you have 6.0 M HCl available.

NH2CH2COOH + HCl NH2CH2COOH.HCl

Moles of glycine.HCl in the solution = 0.1 M x 500 mL = 50 mmol = 50 x 10-3 mol     ( 1 mmol = 10-3 mol)

Thus, moles of glycine required = 50 x 10-3 mol

Also, moles of HCl required = 50 x 10-3 mol

Molar mass of glycine = 75.07 g/mol

Hence, the mass of glycine required = 75.07 g/mol x 50 x 10-3 mol

= 3.75 g

Given, concentration of HCl = 6.0 M

Hence, the volume of HCl required = 50 x 10-3 mol/6.0 M

= 8.3 x 10-3 L

= 8.3 mL          ( 1 L = 1000 mL)

Hence, the volume of water to be added to make the solution = (500 - 8.3) mL = 491.7 mL

Hence, to make the solution, 3.75 g of glycine is to be dissolved in 491.7 mL of water and then 8.3 mL of 6.0 M of HCl is to be added.

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