pue a titration using standard NaOH and HPO, It took 25.00 ml of o.1234NM NaOH t

Question

pue a titration using standard NaOH and HPO, It took 25.00 ml of o.1234NM NaOH to neutralize the H POs using obtained. Bromcresol green indicator. A light GREEN solution was The balanced chemical equation for the reaction is shown below What is the concentration of M,PO used in the experiment? A student performed a titration using standard NaOH and H,PO It took 30.00 mL of 0.1234 M NaOH to neutralize HiPOs using phenolphthalein indicator, A light PINK solution was obtained 4. The balanced chemical equation for the reaction is shown below What is the concentration of HiPO, used in the experiment? 5. A student performed a titration experiment. He put 20.00 ml NaOH in a flask. it took 45,70 mL of 0.5000 M H,SO, to neutralize the NaOH. The balanced chemical equation for the reaction is shown below What is the concentration of NaOH solution used in the experiment?

Explanation / Answer

Ans. #3. Balanced reaction: H3PO4(aq) + 3 NaOH (aq) ------> Na3PO4(aq) + 3 H2O

At equivalence point, the total number of moles of H+ from acid is equal to total number of OH- from base.

That is-           x(M1V1), acid = y(M2V2), base

            Where, x = moles of H+ produced per mol acid = 3 for H3PO4

                        y = moles of OH- produced per mol base = 1 for NaOH

                        V and M are volume and molarity of respective solution.

Note: The volume of H3PO4 solution being titrated is NOT mentioned in the question- without which, the calculation of molarity of acid won’t be possible. So, the calculation is shown using a hypothetical volume of 25.0 mL. Use the correct volume give to you to get the answer.

Let the volume of H3PO4 being titrated be 25.0 mL.

Putting the values in above expression-

                        3 (M1 x 25.0 mL) = 1 x (0.1234 M x 25.00 mL)

                        Or, M1 = (0.1234 M x 25.00 mL) / (3 x 25.0 mL)

                        Hence, M1 = 0.0411 M

Therefore, concentration (molarity) of H3PO4 = 0.0411 M

#4. Balanced reaction: H3PO4(aq) + 3 NaOH (aq) ------> Na3PO4(aq) + 3 H2O

Let the volume of H3PO4 being titrated be 25.0 mL.

Putting the values in above expression-

                        3 (M1 x 25.0 mL) = 1 x (0.1234 M x 10.00 mL)

                        Or, M1 = (0.1234 M x 10.00 mL) / (3 x 25.0 mL)

                        Hence, M1 = 0.01645 M

Therefore, concentration (molarity) of H3PO4 = 0.0165 M

#5. Balanced reaction: H2SO4(aq) + 2 NaOH (aq) ------> Na2SO4(aq) + 2 H2O

          x(M1V1), acid = y(M2V2), base

            Where, x = moles of H+ produced per mol acid = 2 for H2SO4

Putting the values in above expression-

                        2 (0.5000 x 45.70 mL) = 1 x (M2 x 20.00 mL)

                        Or, M2 = 2 (0.5000 x 45.70 mL) / 20.0 mL

                        Hence, M1 = 2.2850 M

Therefore, concentration (molarity) of NaOH = 2.2850 M

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