Question

Calculate the sodium acetate molar concentration of each dilution. Record the calculated values in Data Table 2. The concentration of the 4.7 pH buffer is 0.05 M sodium acetate. Concentration(start) x Volume(start) = Concentration(final) x Volume(final)

		pH 4.7 buffer (10 mL sodium acetate, 10 mL water)	1st dilution (5 mL pH 4.7 buffer solution, 15 mL water)	2nd dilution (5mL of 1st dilution, 15mL water)	3rd dilution (5mL of 2nd dilution, 15mL water)
	Calculation Help	5mL if 0.1M sodium acetate diluted to 10mL	5mL of previous diluted to 20mL (5ml buffer +15 H2O)	5mL of 1st dilution to 20mL (5ml 1st dilution + 15 H2O)	5mL of 2nd dilution to 20mL (5ml 2nd dilution +15 H2O)
	**Concentration of sodium acetate	.05	???	???	???

**(pH=4.7 buffer used 5mL of 0.1M sodium acetate diluted to a total of 10mL from the addition of acetic acid. Use M₁V₁=M₂V₂ to determine the sodium acetate concentration in the buffer. Then continue to use M₁V₁=M₂V₂ to determine the sodium acetate in each dilution.)

Answer 1

Sol . As M1 × V1 = M2 × V2  

where M1 and M2 are initial and final molarities respectively

and V1 and V2 are initial and final volumes respectively

Ist dilution :  

M1 = 0.05 M , V1 = 5 mL , V2 = 20 mL

So , M2 = ( M1 × V1 ) / V2 = ( 0.05 × 5 ) / 20

= 0.0125 M   

2nd dilution :

M1 = 0.0125 M , V1 = 5 mL , V2 = 20 mL

So , M2 = ( M1 × V1 ) / V2 = ( 0.0125 × 5 ) / 20

=  0.003125 M    

3rd dilution :

M1 = 0.003125 M , V1 = 5 mL , V2 = 20 mL  

So , M2 = ( M1 × V1 ) / V2 = ( 0.003125 × 5 ) / 20

=  0.00078125 M