Question

230 mL of 0.10 M NaOH are added to 100 mL of 0.10 M H3PO4. What is the pH of the resulting solution?

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NaOH taken = 0.1 x 230 x 10-3

moles of NaOH taken = 23 x 10-3

also

moles of H3P04 = 0.1 x 100 x 10-3 = 10 x 10-3

now

the reaction is

3NaOH + H3P04 ---> Na3P04 + 3H20

now

we can see that

moles of H3P04 reacted = (1/3) x moles of NaOH taken

so

moles of H3P04 reacted = 23 x 10-3 / 3

moles of H3P04 reacted = 7.667 x 10-3

now

moles of H3P04 left = 10 x 10-3 - 7.667 x 10-3

moles of H3P04 left = 2.333 x 10-3

now

all the NaOH is consumed

and

Na3P04 formed is a nuetral salt

so

the pH is contributed by H3P04 only

now

final volume = 230 + 100 = 330 ml

now

finally

concentration = moles / volume (L)

so

[H3P04] = 2.333 x 10-3 / 330 x 10-3

[H3P04] = 7.07 x 10-3

now

H3P04 ---> 3H+ + P043-

so

[H+] = 3 x [H3P04]

[H+] = 3 x 7.07 x 10-3

[H+] = 0.0212

now

we know that

pH = -log [H+]

so

pH = -log 0.0212

pH = 1.673

so

the pH of the resulting solution is 1.673