Answer to Question #101894 in General Chemistry for Katherine Ewing

Question

Using a stock titrant concentration of 0.1000 mol/L NaOH, determine the NaOH concentrations prepared using the following dilutions: 1:25 dilution (10 mL pipetted in 250 mL volumetric flask) 1:5 dilution (50 mL pipetted in 250 mL volumetric flask) During the quick titration of 10 mL of lemon juice with 0.1053 mol/L NaOH, a student dispensed 82 mL sodium hydroxide (titrant) into the lemon juice (analyte) before the end point was reached. Estimate the analyte concentration, assuming that the acid in the lemon juice is monoprotic. What should be diluted (titrant or analyte) in the titration described in Question 2 above? What dilution factor is necessary so that between 15 and 40 mL of titrant is used in the quantitative titration? See the procedure, pages 0-8 and 0-9 for possible dilutions.

Answer 1

C(NaOH) = 0.1000 mol/L

Solution:

1:25 V(NaOH) = 250 + 10=260 ml=0.26 L

n(NaOH)= 0.1mol/L*0.01L= 0.001 mol

C(NaOH) = 0.001mol/0.26L = 0.0038 mol/L

1:5 V(NaOH) = 250+50=300ml= 0.3L

n(NaOH) = 0.1 mol/L*0.05 L= 0.005 mol

C(NaOH) = 0.005 mol/0.3 L= 0.0166 mol/L

V(C6H8O7) = 10 ml

C(NaOH)= 0.1053 mol/L

V(NaOH)= 82 ml

C6H8O7 + 4NaOH -> C6H4O7Na4 + 4H2O

n(NaOH) = 0.1053 mol/L*0.082L=0.086 mol

n(C6H8O7) = 0.021 mol

C(C6H8O7) = 0.021mol/0.01 L = 0.21 mol/L