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?

~2.054M

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:)

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.