A 25.0 mL sample 0.969 M NaF is titrated 0.220 M HCl standard solution. Hydrochl
ID: 897581 • Letter: A
Question
A 25.0 mL sample 0.969 M NaF is titrated 0.220 M HCl standard solution. Hydrochloric acid is the titrant and sodium fluoride is the analyte. 1. Determine the initial pH of the analyte solution. 2. Determine the pH at the equivalence point. 3. Determine the pH of the analyte solution halfway to the equivalence point. (For example, if 100 mL is required to reach the equivalence point, then calculate the pH of the analyte solution once 50 mL of the titrant has been added.) 4. Determine the pH of the analyte solution 15 mL past the equivalence point. 5. Draw the titration curve based on your calculations. A 25.0 mL sample 0.969 M NaF is titrated 0.220 M HCl standard solution. Hydrochloric acid is the titrant and sodium fluoride is the analyte. 1. Determine the initial pH of the analyte solution. 2. Determine the pH at the equivalence point. 3. Determine the pH of the analyte solution halfway to the equivalence point. (For example, if 100 mL is required to reach the equivalence point, then calculate the pH of the analyte solution once 50 mL of the titrant has been added.) 4. Determine the pH of the analyte solution 15 mL past the equivalence point. 5. Draw the titration curve based on your calculations.Explanation / Answer
NaF is titrated with HCl.
NaF is a salt of weakacid,strongbase.so that it is basic in nature.
a) initial pH of NaF = 7+1/2(pka+log C)
pka of HF = 3.17
C = concentration of salt = (25/1000*0.969) = 0.0242 Mole
pH = 7+1/2(3.17+log0.0242) = 7.77
b) pH at equivalence point depends up on only HCl added
at equivalence point.
No of moles of NaF = No of moles of HCl = 0.0242 Mole
volume of HCl added = 0.0242/0.22 = 0.11 L = 110 ml
concentration of HCl = 0.0242/(0.025+0.11) = 0.179 M
pH = -log(0.179) = 0.7471
c) pH at half equivalence point = pka of HF
pka of HF = 3.17
D) 15 ml of HCl added to equivalence point.at this condition pH depends upon
only excessHCl
concentration of excess HCl = (11/(25+110+15)*0.22) = 0.01613 M
pH = -log(0.01613) = 1.79
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