H 2 SO 4 (aq)

advertisement
KEY
Sample Questions for Chapter 14: Acids and Bases
14.7 Reactions of Acids and Bases (Read pgs. 465- 468 in the chemistry textbook)
14.8 Acid-Base Titration (Read pgs. 468 - 470 in the chemistry textbook)
1.
(a) What is the definition of a salt?
an ionic compound that does not have H+ as the cation or OH- as
the anion.
(b) Here are some examples of making salts:
(i) from acids: take away the H+ and replace it with a cation
Acid
Salt
Formula:
Formula:
H2SO4 = sulfuric acid
Na2SO4 = sodium sulfate
HBr = hydrobromic acid
NaBr
CH3COOH = acetic acid
NaCH3COO = sodium acetate
= sodium bromide
(ii) from bases: take away the OH─ and replace it with an anion
Base
Formula:
Salt
Formula:
Cu(OH)2 = copper (II) hydroxide
CuSO4
= copper (II) sulfate
NaOH
NaBr
= sodium bromide
2.
= sodium hydroxide
(a) What is a neutralization reactions:
a reaction between an acid and a base
(b) What 2 products are formed during a neutralization reaction?
1) water 2) salt
1
3. Write the following equations showing the formation of a salt from the reaction of a metal with and an acid:
(a) aqueous sodium hydroxide + hydrochloric acid → water + aqueous sodium chloride
salt
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
(b) nitric acid and aqueous barium hydroxide → water + aqueous barium nitrate
salt
2 HNO3(aq) + Ba(OH)2(aq) → H2O(l) + Ba(NO3)2(aq)
(c) solid magnesium hydroxide + hydrobromic acid → aqueous magnesium bromide + water
salt
Mg(OH)2 (s) + 2 HBr(aq) → MgBr2 (aq) + 2 H2O(l)
(d) solid zinc hydroxide l + nitric acid → aqueous zinc nitrate + water
salt
Zn(OH)2 (s) + 2 HNO3(aq) → Zn(NO3)2(aq) + 2 H2O(l)
4. Predict the products of the following neutralization reactions. Use a solubility chart to determine the phase
of matter of the salt. Balance the resulting equation.
2 H2O(l) + MgSO4(aq)
(a) H2SO4(aq) + Mg(OH)2(s) →
(b)
2 H3PO4(aq)
(c)
2 NaOH(s)
+
3 Ca(OH)2(s)
→
6 H2O(l) + Ca3(PO4)2 (s)
+ H2SO4(aq) →
2 H2O(l) + Na2SO4(aq)
(d) HNO3(aq) + NH4OH(aq) →
H2O(l) + NH4NO3(aq)
(e)
2 HCl(aq)
+ Pb(OH)2(s) →
(f) Cu(OH)2(s) +
2 HClO4(aq)
2 H2O(l) + PbCl2 (s)
→
2 H2O(l) + Cu(ClO4)2(aq)
2
5. Write complete, balanced equations for each of the following neutralization reactions. Use a solubility chart
to help determine the phase of matter of the salt.
(a) solid lithium hydroxide + phosphoric acid
H3PO4(aq) + 3 LiOH (s) → 3 H2O(l) + Li3PO4 (aq)
(b) hydroioidic acid + aqueous ammonium hydroxide
HI(aq) + NH4OH(aq) → H2O(l) + NH4I(aq)
(c) nitric acid + solid calcium hydroxide
2 HNO3(aq) + Ca(OH)2(s) → 2 H2O(l) + Ca(NO3)(aq)
(c) suluric acid + solid aluminum hydroxide
2 Al(OH)3(s) + 3 H2SO4(aq) → 6 H2O(l) + Al2(SO4)3(aq)
(d) hydrochloric acid + solid iron (III) hydroxide
3 HCl(aq) + Fe(OH)3(s) → 3 H2O(l) + FeCl3 (aq)
(e) sulfuric acid + aqueous barium hydroxide
H2SO4 (aq) + Ba(OH)2(aq) → 2 H2O(l) + BaSO4(s)
ACID / BASE TITRATION
6. Define the following words:
(a) titration
a laboratory procedure in which an acid of unknown
molarity is neutralized with an base of known molarity; or viceversa
6. Continued:
3
(b) titrant
the substance (usually of known molarity) which is used to titrate
the another substance (usually of unknown molairty); the titrant
is always the substance in the buret - regardless of whether or not
its concentration is known
(c) indicator
a substance that dramatically changes color when the pH of the
solution changes
(d) endpoint
the point during a titration when the indicator changes color showing that the solution of unknown molarity has been
neutralized
7.
(a) What is a commonly used indicator?
phenolphthalein
(b) What color is this indicator in an acidic solution?
(c) What color is this indicator in a basic solution?
clear, colorless
pink
8. A few drops of phenolphthalein are added to a 25.0-mL sample of an HCl solution. This solution is then
titrated with a 0.185 M NaOH solution.
(a) Which is the solution of unknown concentration?
(b) What is the indicator?
(c) What is the titrant?
HCl
phenolphthalein
0.185 M NaOH
4
8. Continued:
(d) If 32.6 mL of the titrant were required to neutralize the HCl, what was the molarity of the HCl?
(i) Write a balanced chemical equation to determine the stoichiometry
between the acid and the base:
NaOH + HCl → H2O + NaCl
the stoichiometry is 1:1
it takes 1 mole of NaOH to neturalize 1 mole of HCl
(ii) determine the moles of titrant used
𝟎.πŸπŸ–πŸ“ π’Žπ’π’ 𝑡𝒂𝑢𝑯
𝑳
(0.0326 L )
= 0.00603 moles of NaOH used
= 0.00603 moles of HCl neutralized
(ii) calculate the molarity of the unknown
𝟎.πŸŽπŸŽπŸ”πŸŽπŸ‘π’Žπ’π’π’†π’” 𝑯π‘ͺ𝒍
𝟎.πŸŽπŸπŸ“πŸŽ 𝑳
= 0.241 M HCl
9. What is the molarity of an HCl solution if 28.6 mL of a 0.175 M NaOH solution is needed to neutralize a
25.0-mL sample of the HCl solution?
(i) determine the moles of titrant used
𝟎.πŸπŸ•πŸ“ π’Žπ’π’ 𝑡𝒂𝑢𝑯
𝑳
(0.0286 L )
= 0.00501moles of NaOH used
= 0.00501moles of HCl neutralized
(ii) calculate the molarity of the unknown
𝟎.πŸŽπŸŽπŸ“πŸŽπŸπ’Žπ’π’π’†π’” 𝑯π‘ͺ𝒍
𝟎.πŸŽπŸπŸ“πŸŽ 𝑳
= 0.200M HCl
5
10. What volume, in milliliter, of a 0.115 M NaOH solution would neutralize 25.0 mL of a 0.106 M H2SO4
solution?
(i) Write a balanced chemical equation to determine the stoichiometry
between the acid and the base:
2 NaOH + H2SO4 → H2O + Na2SO4
the stoichiometry is 2:1
it takes 2 moles of NaOH to neturalize 1 mole of H2SO4
(ii) determine the molarity of the acid
𝟎.πŸπŸŽπŸ” π’Žπ’π’ π‘―πŸ π‘Ίπ‘ΆπŸ’
𝑳
(0.0250 L )
= 0.00265 moles of H2SO4
(iii) determine the moles of NaOH needed to neutralize all the H2SO4
0.00265 × 2 = 0.00530 moles of NaOH needed
(iv) calcualte the volume of NaOH needed
𝟎.πŸŽπŸŽπŸ“πŸ‘πŸŽ π’Žπ’π’π’†π’” 𝑡𝒂𝑢𝑯
𝟎.πŸπŸπŸ“ π’Žπ’π’π’†π’” 𝑡𝒂𝑢𝑯
𝑳
= 0.0461 L of NaOH =
46.1 mL NaOH needed
6
11. What volume, in milliliter, of a 0.158 M KOH solution would neutralize 50.0 mL of a 0.212 M HCl
solution?
(i) Write a balanced chemical equation to determine the stoichiometry
between the acid and the base:
KOH + HCl → H2O + KCl
the stoichiometry is 1:1
it takes 1 moles of KOH to neturalize 1 mole of HCl
(ii) determine the molarity of the acid
𝟎.𝟐𝟏𝟐 π’Žπ’π’ 𝑯π‘ͺ𝒍
𝑳
(0.0500 L )
= 0.0106 moles of HCl
= 0.0106 moles KOH needed
(iii) calcualte the volume of NaOH needed
𝟎.πŸŽπŸπŸŽπŸ” π’Žπ’π’π’†π’” 𝑲𝑢𝑯
𝟎.πŸπŸ“πŸ– π’Žπ’π’π’†π’” 𝑲𝑢𝑯
𝑳
= 0.0671 L of NaOH =
67.1 mL KOH needed
7
Download