How many millilitres of 0.120 mol L^-1 NaOH(aq) are required to titrate 50.0 mL of 0.0998 mol L^-1 aqueous butanoic acid to the equivalence point? Butanoic acid is monoprotic.The
Of butanoic acid is 1.5 × 10^-5.

Question

How many millilitres of 0.120 mol L^-1 NaOH(aq) are required to titrate 50.0 mL of 0.0998 mol L^-1 aqueous butanoic acid to the equivalence point? Butanoic acid is monoprotic.The

How many millilitres of 0.120 mol L<sup>-1</sup> NaOH(aq) are required to titrate 50.0 mL of 0.0998 mol L<sup>-1</sup> aqueous butanoic acid to the equivalence point? Butanoic acid is monoprotic.The Of butanoic acid is 1.5 × 10<sup>-5</sup>. A) 4.90 B) 50.0 C) 41.6 D) 60.1 E) 4.65

Of butanoic acid is 1.5 × 10^-5.

Quizplus · Accepted Answer

First, we need to determine how many moles of butanoic acid are present in 50.0 mL of 0.0998 mol/L solution:

0.0500 L x 0.0998 mol/L = 0.00499 mol butanoic acid

Since butanoic acid is monoprotic, it will react with one mole of NaOH according to the balanced equation:

CH3CH2CH2COOH + NaOH → CH3CH2CH2COO- Na+ + H2O

Thus, we need 0.00499 mol of NaOH to react completely with the butanoic acid. To calculate the volume of 0.120 mol/L NaOH required, we can use the following formula:

Volume (L) = moles / concentration (mol/L)

Volume = 0.00499 mol / 0.120 mol/L = 0.0416 L = 41.6 mL

Therefore, the answer is C: 41.6.

How Many Millilitres of 0