Question

Student weighs out 0.287 g of ascorbic acid (H2CH06), a diprotic acid, into a 250. mL k and dilutes to the mark with distilled water. He plans to titrate the acid with tion olume of NaOH Ssolution the student will need to add to reach the final equ ur answer to 3 significant digits. ence I Don't Know Submit | Privacy Oation. All Rights Reserved. Terms of Use ZO19 Micorw

Answer 1

Explanation:

It is given that mass of ascorbic acid = 0.287 g

As molar mass of ascorbic acid is 176.12 g/mol. Hence, calculate its number of moles as follows.

moles ascorbic acid = $\frac{\text{mass ascorbic acid}}{\text{molar mass ascorbic acid}}$

           moles ascorbic acid = $\frac{0.287 g}{176.12 g/mol}$

                                             = $1.63 \times 10^{-3}$ mol

So, moles of $H^{+}$ = $2 \times \text{(moles ascorbic acid)}$

                                 = $2 \times (1.63 \times 10^{-3} mol)$

                                 = $3.26 \times 10^{-3}$ mol

Therefore, moles NaOH required = moles $H^{+}$

     moles NaOH required = $3.26 \times 10^{-3}$ mol

Now, we will calculate the volume of NaOH as follows.

          volume NaOH = $\frac{\text{moles NaOH required}}{\text{molarity NaOH}}$

            volume NaOH = $\frac{3.26 \times 10^{-3} mol}{0.0500 M)}$

                                    = 0.0652 L

or,                                = 65.2 mL              (as 1 L = 1000 ml)

Thus, we can conclude that the volume of NaOH required is 65.2 ml.