Hi again!
I have a new question, Can you help me?
Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000?
My Calculations:
To calculate the concentration of x, I take the pH value ->
pH = 4,000=> x=[H+]= 1.E-4=0.0001M=
x=1.E-4M
The equilibrium reaction is.
HA H+ + A-
Initial x ~0 0
Final x-1.E-4 x-1.E-4 x-1.E-4
Ka=(1,00.E-4)^2/(x-1*E-4)=1,00.E-6;
solving x = 0,0101M
To calculate the quantity of moles in the solution, I do the following:
(50,0*10-3L)* (0,0101M) = 5,05*10-4 mol HA.
I don´t know what to do now!
Thanks!
For a pH = 4, [H+] = antilog(-4.000) = 1.00x10^-4
Ka = [H+][A-] / [HA]
Let [H+] = [A-] = x
Ka = x^2 / (c - x) , where c = molar concentration of HA before dissociation. if c is much larger than x, we can simplify the expression to:
Ka = x^2 / c
(1.00x10^-6)c = 1.00x10^-8
c = 1.00x10^-2 = 0.0100 M
If we want a pH of 5, we can calculate a new value for c (as we did above) which is lower than the previous value of 0.0100 M.
Calculate c for pH = 5, then use:
(0.0100M)(50.0mL) = (c)(V)
substitute the 2nd value of c and solve for V. Then volume added =
V - 50.0mL = ______?
but the rule of 5%?
0,0100/(1.00x10^-4) =100%
The aproximation you do is not valid...or?
Ka= (1,0*10^(-4))^2/(x-1*10^(-4) )=1,00*10^(-6)....I got c=0,0101M
:-(
where did you get c = 1.00x10^-2 = 0.0100 M from ?
To calculate the volume of water that needs to be added to the solution to change the pH from 4.000 to 5.000, you can use the concept of dilution.
First, calculate the moles of HA present in the initial solution by multiplying the volume of the solution (50.0 mL) by the concentration of HA (0.0101 M). This gives you 5.05 x 10^(-4) moles of HA.
To dilute the solution to a pH of 5.000, you can assume that the concentration of HA remains the same. This means that the moles of HA will also remain the same after dilution.
Now, you can use the formula for dilution: C1V1 = C2V2, where C1 and V1 represent the initial concentration and volume, and C2 and V2 represent the final concentration and volume.
You know the initial concentration (0.0101 M) and volume (50.0 mL), and you want to find the final volume (V2) of the solution. The final concentration (C2) is the same as the initial concentration, so you can rewrite the equation as: (0.0101 M)(50.0 mL) = (0.0101 M)(V2).
Simplifying the equation, you get: 0.0101 mol = 0.0101 M * V2.
Canceling out the units, you have: 0.0101 mol = V2 L.
Therefore, the volume of water that needs to be added to the solution is 0.0101 L, or 10.1 mL.
So, to make the pH of the solution change from 4.000 to 5.000, you need to add 10.1 mL of water.