Question

What is the molarity of 15.0 milliliters of an unknown acid that is titrated

with a 1.25 M NaOH solution and required 24.65 milliliters of base to
complete the titration?

Answer 1

~2.054M

Step-by-step explanation:

In this question ,I am going to assume that the acid is monoprotic acid(contains only one hydrogen) which I'll represent by HA.

Chemical equation:

NaOH(aq) + HA(aq)----->NaA(aq) + H2O(l)

The mole ratio of NaOH:HA is 1:1

1.25M NaOH=1.25 Moles/L or 1.25 moles /1000ml of NaOH

(1.25 moles /1000ml) x (24.65ml)=0.0308125 moles NaOH

Mole ratio is 1:1

So, moles of HA are also 0.0308125 moles

volume of HA =15ml

To find molarity of HA:

=(0.0308125 moles x 1000ml)/(15ml)

~2.054M of HA

~Hope it helps:)