A 25.0 ml sample of a solution of a monoprotic acid is titrated with a 0.115 m naoh solution. The titration curve above was obtained. The concentration of the monoprotic acid is about ________ mol/l.

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rejkjavik rejkjavik · Answer 1 · 2019-11-06T05:33:18+01:00

Answer:

concentration of the monoprotic acid is about 0.115 mol/l.

Explanation:

Given data:

volume of solution of monoprotic acid is 25.0 ml

Number of Mole of NaOH in 25.0 ml = 0.115 M

Moles of Monoprotic acid is calculated as

Moles of monoprotic acid [tex]= \frac{0.25L}{0.115 M} = 0.002875 moles[/tex]

Concentration of monoprotic acid is calculated as

Concentration [tex]= \frac{0.002875}{0.025} = 0.115 mol /L[/tex]

AFOKE88 AFOKE88 · Answer 2 · 2021-11-12T07:46:28+01:00

The Concentration of the monoprotic acid is about; 0.23 M

We are given;

Volume of Monoprotic acid = 25 mL

Molarity of NaOH = 0.115 M

Now, a monoprotic acid is defined as an acid that can contribute one proton to a neutralization reaction.

Let the monoprotic acid be identified as HA. Thus, the balanced chemical equation that describes this neutralization reaction would be;

HA(aq) + NaOH(aq) = NA(aq) + H20 (l)

From the titration curve attached, we see that the volume of NaOH added at the Equivalence point is 50 mL.

Thus, since acid volume is 25 mL, then the volume of acid is smaller by a factor of;

50/25 = 2

Thus;

Concentration of the monoprotic acid = 2 × 0.115

Concentration of the monoprotic acid = 0.23 M

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