promotion image of download ymail app
Promoted

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!

Update:

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

1 Answer

Relevance
  • 8 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.

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.