Titration pH 1?

Question

I've only got 5 tries left on this problem. Can anybody help me?

Consider the titration of 50.0 mL of 0.200 M HClO4 by 0.100 M NaOH.

Calculate the pH after 34.7 mL of NaOH has been added.

pH =

At what volume (in mL) of NaOH added does the pH of the resulting solution equal 7.00? Include the units of mL in your answer.

NaOH =

Answer 1

For pH 7, the amounts of H+ and OH- must be equal.

So 50.0 mL HClO4 x 0.200 mol/L HClO4 x 1 H+ per HClO4 must be equal to
V mL NaOH x 0.100 mol/L NaOH x 1 OH- per NaOH.

Some people like to use a formula

n1 C1 V1 = n2 C2V2

for calculations like this.

I leave you to do the arithmetic.

The pH after 34.7 mL NaOH has been added is considerably trickier. What you need to do is this:

Amount of HClO4 initially present is 50.0 mL x 0.200 mol/L = 10 millimoles.

The NaOH neutralises 3.47 millimoles (why?)

So you can work out how many millimoles are left.

That much HClO4 is now dissolved in 84.7 mL (why?)

So use the definition

Concentration (M) = amount (mol) divided by volume (L) = amount (mmol) divided by volume (mL) to find [H+] and then use the definition of pH:

pH = - log [H+]

Good luck!