A Na2CO3(MM = 105.99 G/MOL) standard solution is prepared by transferring 2.4814 grams of primary standard-grade...
A Na2CO3(MM = 105.99 G/MOL) standard solution is prepared by transferring 2.4814 grams of primary standard-grade sodium carbonate to 250.0mL volumetric flask, dissolving the sample in about 100 mL of distilled deionized water and diluted to the mark. A 25.00mL aliquot is taken and titrated with 40.18mL of HCl solution. Calculate the concentration of the HCl solution.
Homework Answers
molarity of Na2CO3 = W*1000/G.M.M * volume of solution
= 2.4814*1000/(105.99*250) = 0.09365M
Na2CO3(aq) + 2HCl(aq) ----------------> 2NaCl(aq) + CO2(g) + H2O(l)
1 mole ----------- 2moles
Na2CO3 ------------------------------------ HCl
M1 = 0.09365M ------------------------ M2 =
V1 = 25ml --------------------------------- V2 = 40.18ml
n1 = 1 --------------------------------------- n2 = 2
M1V1/n1 = M2V2/n2
M2 = M1v1n2/n1V2
= 0.09365*25*2/(1*40.18) = 0.1165M >>>>answer
The concentration of HCl solution =0.1165M >>>>answer
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