Prior to Class Calculations: Calculate the volume and mass required to make 100

Question

Prior to Class Calculations: Calculate the volume and mass required to make 100 mL of 0.10 M solutions of your assigned acid and conjugate base - see page 4 for concentrations and molar masses. Identity of Assigned Acid/Base Conjugate Pair Concentration of Stock Solution of Acid (or Base in case of Ammonia) Actual quantity of acid used:_ Actual quantity of base used: Describe the procedure you used for making the solutions for both the acid and the base. Include relevant experimental details like the glassware used. Show the calculation for determining the actual concentrations of solution made with the solid. Acid Base Actual Concentration Measured ph

Explanation / Answer

a) We have 1.0 M HCl – NaCl (58.44 g/mol)

So calculate volume of 1.0 M HCl to make 100 mL of 0.1 M HCl solution

M1 =1.0 M M2 =0.1 M & V2 =100 mL so calculate V1

so we know M1V1 = M2 V2 i.e. = 1.0 V1 = 0.1 x100

V1 =10 mL

So 10 mL volumerequired to make 100 mL of 0.1 M HCl solution

similarly calculate for NaCl molar mass = 58.44 g.mol

we have prepare 100 mL ( 0.100 L ) of 0.1 M NaCl solution

i.e 0.1 mole per 0.1 L solution of NaCl

we know n = CV= 0.1 mol/L x 0.100 L 0.01 mole of NaCl

Weight in g = n x molar masss = 0.01 x 58.44 =0.5844 g

So 0.5844 g of NaCl is required to make 100 mL of 0.1 M NaCl solution

b) We have 1.0 M Formic acid – Sodium formate (68.01 g/mol)

So calculate volume of 1.0 M Formic acid to make 100 mL of 0.1 M Formic acid solution

M1 =1.0 M M2 =0.1 M & V2 =100 mL so calculate V1

so we know M1V1 = M2 V2 i.e. = 1.0 V1 = 0.1 x100

V1 =10 mL

So 10 mL volumerequired to make 100 mL of 0.1 M Formic acid solution

similarly calculate for Sodium formate molar mass = 68.01 g.mol

we have prepare 100 mL ( 0.100 L ) of 0.1 M Sodium formate solution

i.e 0.1 mole per 0.1 L solution of Sodium formate

we know n = CV= 0.1 mol/L x 0.100 L 0.01 mole of Sodium formate

Weight in g = n x molar masss = 0.01 x 68.01 =0.6801 g

So 0.6801 g of Sodium formate is required to make 100 mL of 0.1 M Sodium formate solution

c) We have 1.0 M Propionic acid – Sodium propaonate (96.06 g/mol)

So calculate volume of 1.0 M Propionic acid to make 100 mL of 0.1 M Propionic acid solution

M1 =1.0 M M2 =0.1 M & V2 =100 mL so calculate V1

so we know M1V1 = M2 V2 i.e. = 1.0 V1 = 0.1 x100

V1 =10 mL

So 10 mL volumerequired to make 100 mL of 0.1 M Propionic acid solution

similarly calculate for Sodium propaonate molar mass = 96.06 g.mol

we have prepare 100 mL ( 0.100 L ) of 0.1 M Sodium propaonate solution

i.e 0.1 mole per 0.1 L solution of Sodium propaonate

we know n = CV= 0.1 mol/L x 0.100 L 0.01 mole of Sodium propaonate

Weight in g = n x molar masss = 0.01 x 96.06 =0.9606 g

So 0.9606 g of Sodium propaonate is required to make 100 mL 0of 0.1 M Sodium propaonate solution

d) We have 1.0 M Ammonia – Ammonium chloride(53.49 g/mol)

So calculate volume of 1.0 M Ammonia to make 100 mL of 0.1 M Ammonia solution

M1 =1.0 M M2 =0.1 M & V2 =100 mL so calculate V1

so we know M1V1 = M2 V2 i.e. = 1.0 V1 = 0.1 x100

V1 =10 mL

So 10 mL volumerequired to make 100 mL of 0.1 M Ammonia solution

similarly calculate for Ammonium chloridemolar mass = 53.49 g.mol

we have prepare 100 mL ( 0.100 L ) of 0.1 M Ammonium chloride solution

i.e 0.1 mole per 0.1 L solution of Ammonium chloride

we know n = CV= 0.1 mol/L x 0.100 L 0.01 mole of Ammonium chloride

Weight in g = n x molar masss = 0.01 x 53.49 =0.5349 g

So 0.5349 g of Ammonium chlorideis required to make 100 mL 0of 0.1 M Ammonium chloride solution

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