# Chemistry Acid/Base Titration Question help?!?

6) what is ph of solution after 25 mL of 0..1 M NaOH is titrated with 30 mL of 0.1 M HCL

a. 2.05

b. 7

c. 5.35

d. 2.6

e. 3.3

Please show working out!

A friend told me it's e but I don't know how she did it

### 1 Answer

- ChemTeamLv 78 months ago
We know that NaOH and HCl react in a 1:1 molar ratio.

We know that the molarities are equal.

Therefore, equal volumes will neutralize each other exactly.

That accounts for 25 mL. What about the left over 5 mL of HCl solution?

Compute the new molarity:

M1V1 = M2V2

(0.1) (5) + (M2) (55)

M2 = 0.009090909090909 M

Now, calculate the pH (knowing that HCl is a strong acid):

pH = -log 0.009090909090909

pH = 2.04

answer choice a

If the molarities were different, we would have to calculate the moles of each reactant (rather than using molarities directly). We would then the amount of left over moles, calculate a new molarity (using moles divided by liters of solution) and then calculate a pH.

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