FIRST U should find whether there is reaction when they are mixed,and then what is the concentration of the mixture.next,find the concentration of H+ ion,finally,use this formular to solve your problem
pH =-log[concentration of H+ ]
Find the pH of mixture of acids.
0.185 M in HCHO2 and 0.225 M in HC2H3O2
Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get the right answer help please?
Answer is pH of 2.19.
4 answers
It you cannot determine the concentration of the mixture by forming a double replacement became you form products that are the same as the reacting because you are mixing two acids.
You must recognize that this is a mixture of two weak acids; i.e., formic acid and acetic acid.
I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH.
Calculate the H^+ from the strong acid, then add the H^+ from the weak acid.
Formic acid first since it is the stronger.
.........HCOOH ==> H^+ + HCOO^-
initial..0.185.....0......0
change...-x........x......x
equil..0.185-x.....x......x
Ka = (H^+)(HCOO^-)/(HCOOH)
Solve for H^+. This is what acts as the common ion (remember Le Chatelier's Principle). This causes the acetic acid to ionize less than it would if were just acetic acid solution.
........CH3COOH ==> H^+ + CH3COO^-
initial..0.225.......0......0
change....-x........x........x
equil...0.225-x......x........x
Ka acet acid = (H^+)(CH3COO^-)/(CH3COOH)
Substitute TOTAL H^+ into the Ka expression for CH3COOH. That will be about 0.00572 from HCOOH from the above calculation plus x from this ionization. Solve for x, add this (H^+) to the 0.00572 from HCOOH, then convert to pH. I obtained pH = 2.19
I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH.
Calculate the H^+ from the strong acid, then add the H^+ from the weak acid.
Formic acid first since it is the stronger.
.........HCOOH ==> H^+ + HCOO^-
initial..0.185.....0......0
change...-x........x......x
equil..0.185-x.....x......x
Ka = (H^+)(HCOO^-)/(HCOOH)
Solve for H^+. This is what acts as the common ion (remember Le Chatelier's Principle). This causes the acetic acid to ionize less than it would if were just acetic acid solution.
........CH3COOH ==> H^+ + CH3COO^-
initial..0.225.......0......0
change....-x........x........x
equil...0.225-x......x........x
Ka acet acid = (H^+)(CH3COO^-)/(CH3COOH)
Substitute TOTAL H^+ into the Ka expression for CH3COOH. That will be about 0.00572 from HCOOH from the above calculation plus x from this ionization. Solve for x, add this (H^+) to the 0.00572 from HCOOH, then convert to pH. I obtained pH = 2.19
I got concentration of HCOOH rxn to be 5.77X10^-3. And the CH3COOH rxn to have concentration of 2.01X10^-3. I added those together to get 7.78X10-3. What do you mean add to ka expression? I keep getting 1.35 now.
1 answer