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To find the molarity of the [tex]\( HNO_3 \)[/tex] solution, let's go through the problem step-by-step carefully, accounting for the given data and the chemical reaction.
### Step 1: Convert the given volumes to liters
The volume of [tex]\( HNO_3 \)[/tex] is given as:
[tex]\[ 22.5 \, \text{mL} = 0.0225 \, \text{L} \][/tex]
The volume of [tex]\( Ca(OH)_2 \)[/tex] is given as:
[tex]\[ 31.27 \, \text{mL} = 0.03127 \, \text{L} \][/tex]
### Step 2: Calculate the moles of [tex]\( Ca(OH)_2 \)[/tex]
Given the molarity [tex]\( M \)[/tex] of [tex]\( Ca(OH)_2 \)[/tex] is 0.167 M, we use the relationship between moles, molarity, and volume:
[tex]\[ \text{Moles of } Ca(OH)_2 = M \times V \][/tex]
[tex]\[ \text{Moles of } Ca(OH)_2 = 0.167 \, \text{M} \times 0.03127 \, \text{L} \][/tex]
[tex]\[ \text{Moles of } Ca(OH)_2 = 0.00522209 \, \text{moles} \][/tex]
### Step 3: Use the balanced chemical equation to find the moles of [tex]\( HNO_3 \)[/tex]
From the balanced equation:
[tex]\[ 2 HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2 H_2O \][/tex]
The mole ratio between [tex]\( HNO_3 \)[/tex] and [tex]\( Ca(OH)_2 \)[/tex] is 2:1. This means 2 moles of [tex]\( HNO_3 \)[/tex] react with 1 mole of [tex]\( Ca(OH)_2 \)[/tex].
So, the moles of [tex]\( HNO_3 \)[/tex] needed are:
[tex]\[ \text{Moles of } HNO_3 = 2 \times \text{Moles of } Ca(OH)_2 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = 2 \times 0.00522209 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = 0.01044418 \, \text{moles} \][/tex]
### Step 4: Calculate the molarity of the [tex]\( HNO_3 \)[/tex] solution
The molarity [tex]\( M \)[/tex] of a solution is given by:
[tex]\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \][/tex]
The volume of the [tex]\( HNO_3 \)[/tex] solution is [tex]\( 0.0225 \, \text{L} \)[/tex]. Therefore, the molarity is:
[tex]\[ M = \frac{0.01044418 \, \text{moles}}{0.0225 \, \text{L}} \][/tex]
[tex]\[ M = 0.4641857777777778 \, \text{M} \][/tex]
Thus, the molarity of the [tex]\( HNO_3 \)[/tex] solution is approximately [tex]\( 0.464 \, \text{M} \)[/tex].
### Step 1: Convert the given volumes to liters
The volume of [tex]\( HNO_3 \)[/tex] is given as:
[tex]\[ 22.5 \, \text{mL} = 0.0225 \, \text{L} \][/tex]
The volume of [tex]\( Ca(OH)_2 \)[/tex] is given as:
[tex]\[ 31.27 \, \text{mL} = 0.03127 \, \text{L} \][/tex]
### Step 2: Calculate the moles of [tex]\( Ca(OH)_2 \)[/tex]
Given the molarity [tex]\( M \)[/tex] of [tex]\( Ca(OH)_2 \)[/tex] is 0.167 M, we use the relationship between moles, molarity, and volume:
[tex]\[ \text{Moles of } Ca(OH)_2 = M \times V \][/tex]
[tex]\[ \text{Moles of } Ca(OH)_2 = 0.167 \, \text{M} \times 0.03127 \, \text{L} \][/tex]
[tex]\[ \text{Moles of } Ca(OH)_2 = 0.00522209 \, \text{moles} \][/tex]
### Step 3: Use the balanced chemical equation to find the moles of [tex]\( HNO_3 \)[/tex]
From the balanced equation:
[tex]\[ 2 HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2 H_2O \][/tex]
The mole ratio between [tex]\( HNO_3 \)[/tex] and [tex]\( Ca(OH)_2 \)[/tex] is 2:1. This means 2 moles of [tex]\( HNO_3 \)[/tex] react with 1 mole of [tex]\( Ca(OH)_2 \)[/tex].
So, the moles of [tex]\( HNO_3 \)[/tex] needed are:
[tex]\[ \text{Moles of } HNO_3 = 2 \times \text{Moles of } Ca(OH)_2 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = 2 \times 0.00522209 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = 0.01044418 \, \text{moles} \][/tex]
### Step 4: Calculate the molarity of the [tex]\( HNO_3 \)[/tex] solution
The molarity [tex]\( M \)[/tex] of a solution is given by:
[tex]\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \][/tex]
The volume of the [tex]\( HNO_3 \)[/tex] solution is [tex]\( 0.0225 \, \text{L} \)[/tex]. Therefore, the molarity is:
[tex]\[ M = \frac{0.01044418 \, \text{moles}}{0.0225 \, \text{L}} \][/tex]
[tex]\[ M = 0.4641857777777778 \, \text{M} \][/tex]
Thus, the molarity of the [tex]\( HNO_3 \)[/tex] solution is approximately [tex]\( 0.464 \, \text{M} \)[/tex].
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