Mod 10 Extra problems

Extra Practice Problem Solutions SOLUTIONS TO THE EXTRA PRACTICE PROBLEMS FOR MODULE #10 1. H2CO3jsjheacid, because it ...

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Extra Practice Problem Solutions

SOLUTIONS TO THE EXTRA PRACTICE PROBLEMS FOR MODULE #10 1. H2CO3jsjheacid, because it lost a to become HCO3'. That makes water the base, because it accepted an to become HsO^. 2. KOH is the base, because when its ions split apart, the hydroxide ion (OH) accepted a n t o become water. This make CtH^O the acid, since it lost an to become C2H5O", which became a part of the sah. 3. Mg(0H)2 is the base, because when its ions split apart, the hydroxide ion (OH) accepted an H^ to become water. This make HCl the acid, since it lost an H^ to become CI", which became a part of the salt.

4. This is an acid and an ionic base, so it will follow the acid + base -> salt + water formula. The salt comes from the metal ion of the base (Al^^) and the negative ion left over when the acid donates it H^ ions (T). To determine the formula, you switch the charges and drop the signs to make AII3. Thus, the unbalanced equation is: HI + Al(OH)3 ^ AII3 + H2O Balancing gives us: 3HI + AirOH)3 AII3 + 3H7O 5. This is an acid and an ionic base, so it will follow the acid + base -> salt + water formula. The sah comes from the metal ion of the base (Na^) and the negative ion left over when the acid donates it H"^ ions. This acid need to donate three H* ions, and when that is done, P04^" is left behind. To determine the formula, you switch the charges and drop the signs to make Na3P04. Thus, the unbalanced equation is: H3PO4 + NaOH ^ Na3P04 + H2O Balancing gives us: H3PO4 + 3NaOH

NaiPOj + 3H7O

6. Since this involves a covalent base, we cannot use the acid + base —> salt + water formula. Instead, we just have to rely on the definitions of acids and bases. Acids donate H^, and bases accept H^. H3PO4 wants to donate three H^ ions, but the base can accept only one, so it takes three bases to get the job done: H3PO4 + 3CHsN -»

P04^'

+ 3CHfiN^

7. a. Molarity is given by number of moles divided by number of liters. We have both those units, so we just divide them: # moles = 1.11 moles HNO, = 0.74 M Concentration = — # liters 1.5 L — b. In order to get concentration, we must have moles and liters. The problem gives us grams and mL, so we must make two conversions:

Solutions and Tests for Exploring Creation With Chemistry

23.1 gNaOH 1 mole NaOH —X

—— = 0.578 moles NaOH 1 40.0 gNaOH 500.0 «iL 0.001 L X = 0.5000 L 1 1 Hit Now we can calculate molarity: # moles = 0.578^moles NaOH = 1.16 moles NaOH = 1.16 M Concentration = — ^^^^ # liters 0.5000 L L

c. In order to get concentration, we must have moles and liters. The problem gives us grams and mL, so we must make two conversions: 1 moleHjCO X 62.0gILCO3 — — — ^ = 0.234 molesHXO 1 200.0 fflfe 0.001 L X = 0.2000 L 1 littL 14.5gH2CO^

Now we can calculate molarity: #moles 0.234 molesH,CO, molesH,CO, Concentration = — = J = 1.17 = 1.17 M # liters 0.2000 L 8. This is a dilution problem, so we use the dilution equation. Mi is 6.78 M, we need to determine Vi Mz is 1.15 M and V2 = 500.0 mL. M,V, = M,W, (6.78 M)-V, = (1.15 M)- (500.0 mL) 1.15 M • 500.0 mL 6.78 M '''^•^ The chemist must take 84.8 mL of the original solution and dilute it with enough water to make 500.0 mL of solution. 9. This is a dilution problem, so we use the dilution equation. Mi is 5.11 M, we need to determine Vi is 4.5 M and V2 = 50.0 mL.

M2

Extra Practice Problem Solutions

(5.11 M) • V, = (4.5 M)- (50.0 mL) 4.5 M-50.0 mL 5.11 M -'"^ You must take 44 mL of the original solution and dilute it with enough water to make 50.0 mL of solution. 10. Remember, titrations are just stoichiometry problems, so first we have to come up with a balanced chemical equation: HBr + NaOH ^ NaBr + H2O Since the endpoint was reached, we know that there was exactly enough base added to eat up all of the acid. First, then, we calculate how many moles of base were added: 1.14 moles NaOH 0.0450 L — X = 0.0513 moles NaOH 1t

1

We can now use the chemical equation to determine how many moles of acid were present: 0.0513 moleoNaOH 1 mole HBr 1 1 mole NaOH = 0.0513 moles HBr Now that we have the number of moles of acid present, we simply divide by the volume of acid to get concentration: # moles = 0.0513 moles Concentration - —— ^ HBr = 3.42 M # liters 0.0150 L 11. First we have to come up with a balanced chemical equation: 2HC1 + Mg(0H)2

MgCl2 +

2H2O

Since the endpoint was reached, we know that there was exactly enough acid added to eat up all of the base. First, then, we calculate how many moles of acid were added: 1.00 moles HCl 0.00351 1 — X = 0.00351 mLHCl 1t

1

We can now use the chemical equation to determine how many moles of base were present:

Solutions and Tests for Exploring Creation With Chemistry

0.00351 molooHCl 1 moleMg(OH), X—:^ . = 0.00176 molesMg(OH), 1 2 molcsIICl Now that we have the number of moles of base present, we simply divide by the volume of acid to concentration: # moles = 0.00176 moles Mg(OH), = 0.0352 M Concentration = — # liters 0.0500 L