15 ml of 3.0 M NaOH are added to the following: 500.0 ml of pure water
500.0 ml of 0.1 M formic acid
500.0 ml of 0.1 M formate of potassium
500.0 ml of a solution containing 0.1 M formic acid and 0.1 M formate of potassium.
Calculate the difference of pH resulting from the addition in all four cases.
2b- What would be the difference of pH after addition of 20.0 ml of NaOH to 500.0 ml of a solution containing 0.1 M formic acid and 0.1 M formate of potassium?
4 answers
How much of this do you know how to do?
again don't know where to start, really just want a step by step so I can do the rest of my homework
I'll get you started but I don't believe you can't do any of it. I don't know what the problem is asking for when it says "difference in pH". Difference in pH from WHAT?
15 mL of 3M NaOH added to 500 mL.
(NaOH) = 3M x (15/515) = ? (Note: This assumes the volumes are additive.)
Then pOH = (NaOH) = -log(OH^-).
15 mL of 3M NaOH added to 500 mL of 0.1M HCOOH.
NaOH + HCOOH ==> HCOONa + H2O
You start with 15 mL x 3M NaOH = 45 millimols.
Add 500 mL of 0.1M HCOOH = 50 millimols.
........HCOOH + NaOH ==> HCOONa + H2O
I.......50.......0.........0........0
add.............45....................
C.......-45.....-45.......+45......45
E........5........0........45......45
Substitute the E line into the Henderson-Hasselbalch buffer solution equation and solve for pH.
15 mL of 3M NaOH added to 500 mL of 0.1M HCOOK doesn't produce anything so the pH is governed by the NaOH. See part 1a above.
15 mL of 3M NaOH added to 500 mL of mixture 0.1M HCOOH/0.1M HCOOK. This is a buffer to which you have added a strong base. Here is what you have.
mmols NaOH added = 45
mmols HCOOH = 500*0.1 = 50
mmols HCOOK = 500*0.1 = 50
.......HCOOH + OH^- ==> HCOO^- + H2O
I.......50.....0........50.......0
added..........45...............
C......-45....-45.......+45.......+45
E.......5......0........95........45
Use the Henderson-Hasselbalch equation and solve for pH.
15 mL of 3M NaOH added to 500 mL.
(NaOH) = 3M x (15/515) = ? (Note: This assumes the volumes are additive.)
Then pOH = (NaOH) = -log(OH^-).
15 mL of 3M NaOH added to 500 mL of 0.1M HCOOH.
NaOH + HCOOH ==> HCOONa + H2O
You start with 15 mL x 3M NaOH = 45 millimols.
Add 500 mL of 0.1M HCOOH = 50 millimols.
........HCOOH + NaOH ==> HCOONa + H2O
I.......50.......0.........0........0
add.............45....................
C.......-45.....-45.......+45......45
E........5........0........45......45
Substitute the E line into the Henderson-Hasselbalch buffer solution equation and solve for pH.
15 mL of 3M NaOH added to 500 mL of 0.1M HCOOK doesn't produce anything so the pH is governed by the NaOH. See part 1a above.
15 mL of 3M NaOH added to 500 mL of mixture 0.1M HCOOH/0.1M HCOOK. This is a buffer to which you have added a strong base. Here is what you have.
mmols NaOH added = 45
mmols HCOOH = 500*0.1 = 50
mmols HCOOK = 500*0.1 = 50
.......HCOOH + OH^- ==> HCOO^- + H2O
I.......50.....0........50.......0
added..........45...............
C......-45....-45.......+45.......+45
E.......5......0........95........45
Use the Henderson-Hasselbalch equation and solve for pH.
Calculate the pH of a solution that is 0.185M benzoic acid and 0.160M sodium benzoate, a salt whose anion is the conjugate base of benzoic acid. Ka = 6.3 X 10-5
2 answers