Question

What is the ph of a solution containing 10 ml of 1.0 m hcl and 20 ml of 1.0 m naoh?

Answer 1

we know that

HCl is an acid and NaOH is a base

it is common knowledge that an acid will react with a base

so in this case

HCl reacts with NaOH to form NaCl and H20

the reaction is given by

HCl + NaOH --> NaCl + H20

now

we know that

concentration = moles x 1000 / volume (ml)

then

moles = concentration x volume (ml) / 1000

using this definition we can get the moles of HCl and NaOH taken

moles of HCl taken = 1 x 10 / 1000 = 0.01

moles of NaOH taken = 2 x 10 / 1000 = 0.02

now

consider the reaction

NaOH + HCl ---> NaCl + H20

we can clearly see that

1 mole of HCl will react with 1 mole of NaOH
so
0.01 mole of HCl will react with 0.01 mole of NaOH

after 0.01 mole of HCl is consumed , no HCl remains but NaOH is in excess

moles of NaOH unreacted = moles NaOH taken - moles NaOH reacted

moles of NaOH unreacted = 0.02 - 0.01 = 0.01

now

final volume = 10 + 20 = 30 ml

now remember the formula for concentration

concentration = moles x 1000 / volume (ml)

so

concentration of NaOH unreacted = 0.01 x 1000 / 30 = 0.3333

now

we know that NaOH is a very strong base , it undergoes 100% dissociation in water

NaOH ---> Na+ + OH-

we can see that

[OH-] produced = [NaOH] in the solution

[OH-] produced = 0.3333

now

we know that

pOH = -log [OH-]

pOH = -log 0.33333

pOH = 0.477

now

remember that

pH + pOH = pKw = 14

so

pH = 14 - pOH
pH = 14 - 0.477
pH = 13.523

so

pH of the given solution is 13.523
I think the answer can help you very well