Question

LABORATORY EXPERIMENTS IN GENERAL CHEMISTRY 154 If 0,1000 M is used as the molarity of the NaOH solution in question 6, would 7 the final %Na2CO3 in the soda ash sample be greater than, less than, or equal tothetrue % Ña,соз in the soda ash sample? Failing to dry the potassium acid phthalate prior to its use would 8. a) cause the molarity of the NaOH you obtain to be greater than, less than, or equal to the true molarity of the NaOH solution? b) cause the % Na2CO3 in the soda ash sample to be greater than, less than, or equal to the true %Na2CO in the soda ash less than, or Complete and balance the following acid-base reactions: 9. a. KOH + HC,H,02→ b. NaOH + HC,H,O2- c. CsOH-H,so,- d. Ca(OH)2-HF→ 10. If the three titrations of the KHP with the NaOH gave the following molari- ties (0.1372, 0.1445, and 0.1454); then calculate the average molarity, the three deviations, and the average deviation. Is a fourth titration needed (explain)? 11. Another titration was done to go along with the data in question 10. This fourth titration gave a molarity of 0.1437. Calculate the new average molarity, the three new deviations, and the new average deviation. Is a fifth titration needed (explain)?

Answer 1

9. Complete and balance acid-base reactions

Acid-base reactions are these that involves a proton exchange (a proton is a H atom) between two reagents to produce water and a salt. In these reactions, the bases are the reagents capable to catch a proton from an acid. For the reactions given:

a. KOH + HC₂H3O₂

The first H atom in the second reagent will be catched by the base to produce water:

$KOH+HC_2H_3O_2\rightarrow H_2O+K^++C_2H_3O_2^-$

Then, negative and positive ions, will produce the salt. The overall reaction is like a double displacement reaction. It is to say:

$KOH+HC_2H_3O_2\rightarrow H_2O+KC_2H_3O_2$

b. NaOH + HC₇H₅O₂

Following the same procedure, we have:

$NaOH+HC_7H_5O_2\rightarrow H_2O+NaC_7H_5O_2$

c. CsOH + H₂SO₃

To undergo a double displacement reaction, only one hydrogen atom is exchanged here:

$CsOH+H_2SO_3\rightarrow H_2O+CsHSO_3$

d. Ca(OH)₂ + HF

Here, we will need two molecules of HF (the acid) to produce water. Notice that the base has two OH^- ions, and they need two protons to produce two water molecules. Also, to have a neutral salt as a product, we need two F^- ions for every Ca⁺² ion:

Ca(OH)2 + 2HF-2H20+CaF

e. Ba(OH)₂ + H₂SO₄

In this case, the acid provides two hydrogen atoms to produce water, and SO₄^-2 ions reacts with exactly one Ba⁺² ion to produce a salt:

Ba(OH)2 + H2SO4 → 2H20+ BaSO4

f. Al(OH)₃ + H₂TeO₄

Three protons are needed to produce three water molecules. To undergo this reaction in whole numbers, we multiply Al(OH)₃ by 2 and H₂TeO₄ by three, then, we can produce six water molecules, and we have:

$2Al(OH)_3+3H_2TeO_4\rightarrow 6H_2O+Al_2(TeO_4)_3$

g. Sr(OH)₂ + H₃PO₄

From H₃PO₄ we only need two hydrogen atoms to produce two molecules of water, then:

$Sr(OH)_2+H_3PO_4\rightarrow 2H_2O+SrHPO_4$

h. Na₂CO₃ + HCl

In this case, to produce a salt, we need two Cl^- ions to have two molecules of NaCl. If we multiply HCl by two, we have:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3$

And H₂CO₃ inmediately dissociates into CO₂ and water according to:

$H_2CO_3\rightarrow H_2O+CO_2$

Then, the overall reaction is:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2$