Calculating pH Change from HClO4 Addition

[answerseeker]

when you add 10.0 mL of 1.0M HClO4 into 990.0mL water, the pH changes by 2.
how would you find the change in pH?

 

[Gokul43201]

1. Write out the expression that defines pH
2. Assume HClO4 is close to 100% dissociated in solution
3. From the original concentration of HClO4, find [H+] and calculate the original pH from this

4. Calculate the new (diluted) concentration of HClO4 (molarity = number of moles/volume of solution)
5. Using the new value of [H+], calculate pH again.
6. Subtract from the original number to find the difference.

 

[ravenprp]

To find the change in pH, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to its acid concentration and dissociation constant. In this case, we know the initial concentration of HClO4 (1.0M) and the final volume of the solution (1000.0mL). We can also assume that the initial pH of the water was 7, as it is neutral.

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), we can rearrange it to solve for the change in pH:

Change in pH = pH final - pH initial
= (pKa + log([A-]/[HA])) - 7

To find the final concentrations of HClO4 and its conjugate base, we can use the dilution equation: M1V1 = M2V2.

M1 = initial concentration of HClO4 = 1.0M
V1 = initial volume of HClO4 = 10.0mL
V2 = final volume of solution = 1000.0mL
M2 = final concentration of HClO4 = (M1V1)/V2 = (1.0M x 10.0mL)/1000.0mL = 0.01M

Since HClO4 is a strong acid, it will completely dissociate in water, so the final concentration of its conjugate base (ClO4-) will be equal to the final concentration of HClO4 (0.01M).

Now we can plug in these values into the Henderson-Hasselbalch equation:

Change in pH = (pKa + log([0.01M]/[0.01M])) - 7
= (pKa + log(1)) - 7
= pKa - 7

The pKa value for HClO4 is approximately -7, so the change in pH would be:

Change in pH = (-7) - 7 = -14

This means that the pH of the solution would decrease by 14 units when 10.0 mL of 1.0M HClO4 is added to 990.0mL of water. In this case, the pH would decrease from 7 to -7, which is a highly acidic solution. It