fak chem hw 7 - To(st22362 Homework 7 Fakhreddine(52420...

To (st22362) – Homework 7 – Fakhreddine – (52420) 1
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001 1.0 points
We find that 61.1 mL of an HCl solution
reacts exactly with 74.2 mL of 1.371 M KOH
solution. Calculate the molarity of the HCl
solution.
1. 0.832 M
2. 0.667 M
3. 3.33 M
4. 1.17 M
5. 1.66 M correct
Explanation:
V
1
= 61 . 1 mL V
2
= 74 . 2 mL
M
2
= 1 . 371 M
The balanced equation for the reaction is
HCl + KOH → KCl + H
2
O
Molarity is moles solute per liter of solution.
We know the volume of the HCl solution. If
we could find the moles of HCl in the solution
we could calculate the molarity.
We start by calculating the moles of KOH
present:
? mol KOH = 74 . 2 mL soln
×
1 L soln
1000 mL soln
×
1 . 371 mol KOH
1 L soln
= 0 . 1017 mol KOH
Using the mole-to-mole ratio from the bal-
anced chemical equation we calculate the
moles of HCl needed to react with this amount
of KOH:
? mol HCl = 0 . 1017 mol KOH
×
1 mol HCl
1 mol KOH
= 0 . 1017 mol HCl
This is the number of moles of HCl needed to
react with the KOH and therefore the num-
ber of moles present in the 61.1 mL of HCl
solution. We can calculate the molarity of the
HCl solution by dividing the moles of HCl by
the volume of the HCl solution:
? M HCl =
0 . 1017 mol HCl
0 . 0611 L soln
= 1 . 66 M HCl
002 1.0 points
200 mL of 1.5 M HCl is mixed with 150 mL of
3.0 M NaOH.
HCl + NaOH → NaCl + H
2
O
What is the molarity of the resulting salt
solution?
1. 0.86 M NaCl correct
2. 0.45 M NaCl
3. 1.5 M NaCl
4. 2.25 M NaCl
5. 2.0 M NaCl
Explanation:
V
1
= 200 mL M
1
= 1 . 5 M
V
2
= 150 mL M
2
= 3 . 0 M
Our first step is to determine the number of
moles of each reactant present:
? mol HCl = 200 mL soln ×
1 L soln
1000 mL soln
×
1 . 5 mol HCl
1 L soln
= 0 . 3 mol HCl
? mol NaOH = 150 mL soln ×
1 L soln
1000 mL soln
×
3 . 0 mol NaOH
1 L soln
= 0 . 45 mol NaOH
From the balanced chemical equation, we
know that one mole of HCl is needed to re-
act with each mole of NaOH. We don’t have

To (st22362) – Homework 7 – Fakhreddine – (52420) 2
enough HCl to react all the NaOH, so HCl is
our limiting reactant and will determine how
much NaCl is produced:
? mol NaCl = 0 . 3 mol HCl ×
1 mol NaCl
1 mol HCl
= 0 . 3 mol NaCl
We are asked to find the molarity of the salt
solution. Molarity is moles solute per liter of
solution. The total volume of the resulting
solution will be the combined volume of the
two original solutions that were mixed:
? mL NaCl soln
= 200 mL NaCl + 150 mL NaCl
= 350 mL NaCl
We find the molarity of NaCl in the solution
by dividing the moles of NaCl by the volume
of the solution:
Molarity of NaCl =
0 . 3 mol NaCl
350 mL soln
×
1000 mL soln
1 L soln
= 0 . 86 M NaCl
003 1.0 points
Calculate the pH of a solution when 0.200
moles of solid NaOH is added to 2.00 liter of
0.250 M HCl. Assume no change in volume.
1. 1.40
2. 0.82 correct
3. 0.59
4. 1.25
5. 1.53
Explanation:
n
NaOH
= 0 . 200 mol
n
HCl
= (2 . 0 L)(0 . 250 M) = 0 . 5 mol
NaOH + HCl → Na
+
+ Cl
-
+H
2
O
ini 0.200 0 . 5 0 0
Δ − 0 . 200 − 0 . 200 0.200 0.200
fin 0 0 . 3 0.200 0.200
HCl is a strong acid, and Na
+
and Cl
-
are
spectator ions.