Chem119-935am-quiz9-key
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Barondeau Chem 119 9:35 am class Quiz 9
1.
You mix 55 mL of 1.50
M
silver nitrate (AgNO
3
) with 35 mL of 1.47
M
sodium chloride (NaCl).
Determine the mass (with 2 significant figures) of precipitated silver chloride (AgCl; 143.32
g/mol) that would form (2 points).
AgNO
3
(aq) + NaCl(aq)
AgCl(s) + NaNO
3
(aq)
Ag
+
(aq) + NO
3
-
(aq) + Na
+
(aq) + Cl
-
(aq)
AgCl(s) + Na
+
(aq) + NO
3
-
(aq)
Remove spectator ions
Ag
+
(aq) + Cl
-
(aq)
AgCl(s)
Limiting reagent problem
(1.50 moles AgNO
3
/ L)(0.055 L) = 0.0825 moles AgNO
3
or Ag
+
(1.47 moles NaCl / L)(0.035 L) = 0.05145 moles NaCl or Cl
-
Cl
-
is limiting reagent
(0.05145 moles Cl
-
)(1 mol AgCl / 1 mol Cl
-
)(143.32 g AgCl / mol) = 7.3738 or 7.4 g
2.
You have 51.5 mL of a 2.80
M
solution of Na
2
CrO
4
(
aq
). You also have 112 mL of a 2.50
M
solution of AgNO
3
(
aq
). Calculate the concentration of Na
+
after the two solutions are mixed.
Report the answer with 3 significant figures (2 points)
Determine number of moles Na+ and then divide by total volume.
(2.80 moles Na
2
CrO
4
/ L)(0.0515 L)(2 Na
+
/Na
2
CrO
4
) = 0.2884 moles Na
+
Total volume = 51.5 mL + 112 mL = 163.5 mL
Concentration = (0.2884 moles / 0.1635 L) = 1.76 M Na
+
3.
The hydronium ion concentration in a solution was measured to be 4.8
10
-6
M.
a.
What is the pH of this solution?
Report the answer with 2 significant figures (1 point).
pH = -log [H
+
]
pH = -log (4.8
10
-6
M) = 4.7
b.
Is this an acidic or basic solution (1 point)?
At room temperature: pH below 7 is acidic.
pH above 7 is basic.
Acidic
7.4 g
1.76 M
5.3
4.
Write a net ionic equation for the reaction of the weak acid HF with the strong base KOH (2
points)
HF(aq) + KOH(aq)
KF(aq) + H
2
O(
l
)
HF(aq) + K
+
(aq) + OH
-
(aq)
K
+
(aq) + F
-
(aq) + H
2
O(
l
)
Remove spectator ions
HF(aq) + OH
-
(aq)
F
-
(aq) + H
2
O(
l
)
5.
Calculate the mass of NaOH (39.997 g/mol) needed to completely neutralize 58.3 mL of a 0.240
M solution of the monoprotic acid HNO
3
. Report the answer with 3 significant figures (2 points).
NaOH(aq) + HNO
3
(aq)
NaNO
3
(aq) + H
2
O(
l
)
(0.240 moles HNO
3
/ L)(0.0583 L) = 0.013992 moles HNO
3
(0.013992 moles HNO
3
)(1 mol NaOH / 1 mol HNO
3
)(39.997 g NaOH / mol) = 0.559638 g
HF(aq) + OH
-
(aq)
F
-
(aq) + H
2
O(
l
)
0.560 g
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