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?
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.
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
To calculate the difference in pH resulting from the addition of NaOH in each case, we need to consider the reaction that occurs between NaOH and the other substances.
First, let's understand the reaction that takes place when NaOH reacts with water:
NaOH + H2O -> Na+ + OH- + H2O
1. Addition to 500.0 ml of pure water:
Since there are no other substances present except for water, the NaOH will completely dissociate into Na+ and OH-. The concentration of OH- ions will increase, leading to an increase in pH. You can calculate the new pH using the pOH formula and then subtract it from 14 to get the difference in pH.
2. Addition to 500.0 ml of 0.1 M formic acid:
Formic acid (HCOOH) is a weak acid. The reaction that occurs between NaOH and formic acid is a neutralization reaction:
NaOH + HCOOH -> Na+ + HCOO- + H2O
We can use the Henderson-Hasselbalch equation to calculate the pH difference:
pH = pKa + log ([HCOO-] / [HCOOH])
Before the addition, the ratio of formate (HCOO-) to formic acid (HCOOH) is 0.1 M / 0.1 M = 1.
After the addition of NaOH, 15 ml of the 3.0 M NaOH will react with 15 ml of the 0.1 M formic acid, resulting in the formation of 15 ml of 0.1 M formate. Therefore, the new ratio is (0.1 M + 0.1 M) / (0.1 M - 0.1 M) = 1.
3. Addition to 500.0 ml of 0.1 M formate of potassium:
Since formate ions (HCOO-) are already present, the additional NaOH will react with the formate ions to produce more OH-. Similar to the previous case, we can use the Henderson-Hasselbalch equation to calculate the pH difference.
4. Addition to 500.0 ml of a solution containing 0.1 M formic acid and 0.1 M formate of potassium:
In this case, both formic acid and formate ions are present. The reaction will occur between NaOH, formic acid, and formate ions, resulting in the formation of Na+ ions, formate ions, and water. You can use the Henderson-Hasselbalch equation to calculate the pH difference.
To answer the second part of the question:
To calculate the difference in pH after the addition of 20.0 ml of NaOH to 500.0 ml of the solution containing 0.1 M formic acid and 0.1 M formate of potassium, follow the same steps as explained above, but substitute the volumes accordingly.