Solving pH of Solution After NaOH Addition

[MichaelXY]

[SOLVED] pH of solution

Homework Statement

What is the pH of a solution in which 30.0mL of 0.010 M NaOH are added to 10.0mL of .010 M H2SO4?

Homework Equations

The Attempt at a Solution

So first thing I did was write the equation.

2NaOH + H2SO4 ----> Na2SO4 + 2H2O

I never used this equation after this which makes me think I already messed up. I perform the following:

.03L(.010M) = 3*10^-4 mol NaOH
.01L(.010M) = 1*10^-4 mol H2SO4

I then compute concentration of H2SO4 by: 1*10^-4/.01L = .01
pH = -log(.01) = 2.0 pH
I suspect I am missing something here. Any pointers?
Thanks

 

[Gannon]

Everything looks good until the last calculation. It's a strong acid mixed with strong base, so the moles of each will neutralize each other until one runs out. You will then be left with an excess acid or base. However, the concentration of this acid or base is now more dilute because you now have added the solutions together, which adds their respective volumes together. Now you can figure out how this affects the last step. The excess H2SO4 concentration is the number of excess moles divided by total volume, not just by its initial volume.

 

[Blueshift5]

Good job on setting up the equation and calculating the moles of NaOH and H2SO4. However, you are missing the fact that NaOH is a strong base and H2SO4 is a strong acid. This means that they will completely dissociate in water, leading to a neutralization reaction. The products of this reaction will be a salt (Na2SO4) and water (H2O).

To solve for the pH, you need to use the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

Since NaOH is a strong base, it will fully dissociate, leaving only the H2SO4 as the acid. The pKa value for H2SO4 is -3.

pH = -3 + log([1*10^-4]/[3*10^-4])
pH = -3 + log(0.333)
pH = -3 + (-0.477)
pH = -3.477

Therefore, the pH of the solution after the addition of NaOH is 3.477.