A0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 ml of water and titrated with 0.0633 m naoh solution. the volume of base required to bring the solution to the equivalence point was 18.4 ml. (a) calculate the molar mass of the acid. (b) after 10.0 ml of base had been added during the titration, the ph was determined to be 5.87. what is the ka of the unknown acid?

(a) 110 g/mol

(b) Ka = 2.14x10⁻⁵

Explanation:

(a)

The moles of OH⁻ at the equivalence point are equal to the moles of H⁺. Because the acid is monoprotic, the moles of H⁺ are equal to the moles of acid.

Now that we have the moles of acid and its mass, we can calculate its molar mass:

(b)

We can use the Henderson-Hasselbach equation to solve this part:

So we calculate the molar concentration of the basic (deprotonated, A⁻) and the acid (protonated, HA) forms:

The final volume is (25.0 mL + 10.0 mL) = 35.0 mL

[A⁻] ⇒ 10.0mL * 0.0633 M / 35 mL = 0.018 M

[HA] ⇒ (1.16 mmol acid - 10.0mL * 0.0633 M) / 35 mL = 0.015 M

Finally we calculate pKa and from it, Ka:

Ka = = 2.14 x 10⁻⁵

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