3. What is the pH of the resulting solution after 150 mL of a 3.2 M sodium hydro
ID: 548272 • Letter: 3
Question
3. What is the pH of the resulting solution after 150 mL of a 3.2 M sodium hydroxide solution is mixed with 4.0 mL of a 4.0 M vinegar solution?
4. In the above question is the acid or base a limiting reagent?
5. Calculate the pH after adding 40.0 mL of 3.7 M potassium hydroxide to 20.0 mL of 12 M formic acid.
6. How many more mL of the 3.7 M potassium hydroxide solution must be added to the formic acid solution to reach the equivalence point in the above question (#5) above?
7. Calculate the pH of a solution made by adding 70 mL of 4.3 M sodium acetate to 30 mL of 10 M acetic acid.
*Help with any of these could be so wonderful; please show/tell steps*
Explanation / Answer
3)
no of mol of NaoH added = 150*3.2/1000 = 0.48 mol
no of mol of Vinegar(aceticacid) = 4*4/1000 = 0.016 mol
limiitng reactant = vinegar(aceticacid)
pH = pka + log(base/(acid-base))
pka of aceticacid = 4.74
pH = 4.74 + log(0.016/(0.48-0.016))
= 3.28
4) limiitng reactant = vinegar(aceticacid)
5) no of mol of KoH added = 40*3.7/1000 = 0.148 mol
NO of mol of formicacid = 20*12/1000 = 0.24 mol
pH = pka + log(base/acid)
pka of formic acid = 3.75
pH = 3.75+log(0.148/(0.24-0.148))
= 3.96
6) NO of mol of formicacid = 20*12/1000 = 0.24 mol
NO of mol of KOH required to reach equivalence point = 0.24 mol
volume of KOH required = n/M = 0.24/3.7 = 0.065 L
= 65 ml
Volume of extra KOH required = 65 - 40 = 25 ml
7) no of mol of sodium acetate = 70*4.3/1000 = 0.301 mol
no of mol of aceticacid = 30*10/1000 = 0.3 mol
equivalence point
pH = pka +log(sodium acetate/aceticacid)
pH = pka +log(0.3/0.3)
pH = pka = 4.74
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