Carbonate mixtures??? please

Suppose that #0.3000g# sample consist of 50% by weight #"NaOH"# and 50% by weight #"Na"_2"CO"_3#. (a) How many milliliters of #0.1000M# #"HCI"# would be required for titration to the phenolphthalein end point? (b) How many milliliters of #HCl# would have been required had methyl orange indicator been used instead of phenolphthalein?
answers (a,51.65)(b, 65.80)

1 Answer
Ernest Z.
Feb 2, 2018

(a) 51.65 mL; (b) 65.80 mL

Explanation:

What's happening here?

You are titrating a mixture of two bases, #"NaOH"# and #"Na"_2"CO"_3#, with #"HCl"#.

The neutralization of #"Na"_2"CO"_3# takes place in two stages:

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In Stage 1, you are converting the #"CO"_3^"2-"# to #"HCO"_3^"-"#:

#"CO"_3^"2-" + "H"_3"O"^"+" → "HCO"_3^"-" + "H"_2"O"#

The endpoint occurs during the colour change of phenolphthalein.

In Stage 2, you are converting the #"HCO"_3^"-"# to #"H"_2"CO"_3#:

#"HCO"_3^"-" + "H"_3"O"^"+" → "H"_2"CO"_3 + "H"_2"O"#

The endpoint occurs during the colour change of methyl orange.

It takes 2 mol #"HCl"# to neutralize 1 mol #"Na"_2"CO"_3#,

#"NaOH"# is a strong base. One mole of #"HCl"# neutralizes 1 mol #"NaOH"#, and the endpoint occurs during the phenolphthalein colour change.

Thus, at the phenolphthalein endpoint, you have neutralized all the #"HCl"# and half the #"Na"_2"CO"_3#.

At the methyl orange endpoint, you have neutralized the remaining half of the #"Na"_2"CO"_3#.

(a) Titration to phenolphthalein endpoint

(i) Calculate the moles of #"NaOH"# and #"Na"_2"CO"_3#

You have 0.1500 g #"NaOH"# and 0.1500 g of #"Na"_2"CO"_3#.

#"Moles of NaOH" = 0.1500 color(red)(cancel(color(black)("g NaOH"))) × "1 mol NaOH"/(40.00 color(red)(cancel(color(black)("g NaOH")))) = "0.003 750 mol NaOH"#

#"Moles of Na"_2"CO"_3 = 0.1500 color(red)(cancel(color(black)("g Na"_2"CO"_3))) × ("1 mol Na"_2"CO"_3)/(105.99 color(red)(cancel(color(black)("g Na"_2"CO"_3)))) = "0.001 415 mol Na"_2"CO"_3#

Thus, at the phenolphthalein endpoint, you will have neutralized #"0.003 750 mol NaOH"# and converted #"0.001 415 mol CO"_3^"2-"# to #"HCO"_3^"-"#

(ii) Calculate the volume of #"HCl"# required

To neutralize the #"NaOH"#,

#"Volume of HCl" = "0.003 750" color(red)(cancel(color(black)("mol NaOH"))) × (1 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("mol NaOH")))) × "1 L HCl"/(0.1000 color(red)(cancel(color(black)("mol HCl")))) = "0.037 50 L HCl" = "37.50 mL HCl"#

To neutralize the #"CO"_3^"2-"#,

#"Volume of HCl" = "0.001 415" color(red)(cancel(color(black)("mol CO"_3))) × (1 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("mol CO"_3^"2-")))) × "1 L HCl"/(0.1000 color(red)(cancel(color(black)("mol HCl")))) = "0.014 15 L HCl" = "14.15 mL HCl"#

∴ #"Total volume HCl = (37.50 +14.15) mL HCl = 51.65 mL HCl"#

(b)Titration to the methyl orange endpoint

At the methyl orange endpoint, you have converted the #"HCO"_3^"-"# to #"H"_2"CO"_3#.

This requires another 14.50 mL #"HCl"#.

∴ #"Total volume HCl = (51.65 +14.15) mL HCl = 65.80 mL HCl"#

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