Exactly 50.00 mL of an HCl solution required 24.75 mL of 0.02043M Ba(OH)_2 to re

Question

Exactly 50.00 mL of an HCl solution required 24.75 mL of 0.02043M Ba(OH)_2 to reach the end point. Calculate the molarity of the HCl. pKa of H_2CO_3 and HCO_3 are respectively 6.4 and 10.3. Make a rough sketch of the titration curve, which should be obtained upon titration of Na_2CO_3 with HCl. Explain the effect of boiling the solution just before the end point is reached in a titration of carbonate. (Harris pp. 221) Calculate the amount of 0.1000M HCl used to titrate 0.2124g Na_2CO_3. Why is it necessary to titrate the solution until the colour changes from blue to green, and not from blue to yellow?

Explanation / Answer

1)
the reaction that takes place is:
2HCl + Ba(OH)2   <----> BaCl2 + H2O

number of mol of HCl reacting = 2* number of mol of Ba(OH)2
M(HCl)*V(HCl) = 2*M(Ba(OH)2)*V(Ba(OH)2)
M(HCl) * 50.00 = 2*0.02043 * 24.75
M(HCl) = 0.02023 M

Answer: 0.02023 M

i am allowed to answer only 1 question at a time

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