If 300 mL of 3.9 molar HCl solution is added
to 200 mL of 2.2 molar NaOH solution, what
will be the molarity of NaCl in the resulting
solution?
Answer in units of M
Since HCl and NaOH react in a 1:1 ratio, 0.200 L x 2.2 M NaOH = 0.440 mols will react with 0.200 x 3.9 = 1.17 mols HCl to form (ONLY) 0.44 mol NaCl.
M = mols/L soln (careful--the solution volume will be 300 + 200 = 500 mL or 0.5L).
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To determine the molarity of NaCl in the resulting solution, we need to use the concept of moles and the balanced chemical equation for the reaction between HCl and NaOH. The balanced chemical equation is:
HCl + NaOH -> NaCl + H2O
First, let's calculate the number of moles of HCl and NaOH:
1. Moles of HCl = (volume in liters) x (molarity)
Moles of HCl = (300 mL / 1000 mL/L) x 3.9 M
Moles of HCl = 0.3 L x 3.9 M
Moles of HCl = 1.17 moles
2. Moles of NaOH = (volume in liters) x (molarity)
Moles of NaOH = (200 mL / 1000 mL/L) x 2.2 M
Moles of NaOH = 0.2 L x 2.2 M
Moles of NaOH = 0.44 moles
Since the balanced equation shows a 1:1 stoichiometric ratio between HCl and NaCl, the moles of NaCl formed will be the same as the moles of NaOH added.
3. Moles of NaCl = Moles of NaOH
Moles of NaCl = 0.44 moles
Finally, we need to calculate the molarity of NaCl in the resulting solution:
4. Molarity of NaCl = (moles of NaCl) / (volume in liters)
Molarity of NaCl = 0.44 moles / (300 mL + 200 mL) / 1000 mL/L
Molarity of NaCl = 0.44 moles / 0.5 L
Molarity of NaCl = 0.88 M
Therefore, the molarity of NaCl in the resulting solution is 0.88 M.
To find the molarity of NaCl in the resulting solution, we need to use the concept of stoichiometry and the principle of conservation of mass.
Step 1: Determine the moles of HCl and NaOH used
Moles of HCl = Volume of HCl solution (in liters) * Molarity of HCl
Moles of NaOH = Volume of NaOH solution (in liters) * Molarity of NaOH
First, we need to convert the volumes from milliliters to liters:
Volume of HCl solution = 300 mL = 300 / 1000 = 0.3 L
Volume of NaOH solution = 200 mL = 200 / 1000 = 0.2 L
Next, we calculate the moles of HCl and NaOH:
Moles of HCl = 0.3 L * 3.9 M = 1.17 moles
Moles of NaOH = 0.2 L * 2.2 M = 0.44 moles
Step 2: Determine the reaction between HCl and NaOH
HCl + NaOH -> NaCl + H2O
From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl.
Step 3: Determine the moles of NaCl formed
Since the reaction between HCl and NaOH is a 1:1 ratio, the number of moles of NaCl formed is equal to the number of moles of NaOH used.
Moles of NaCl formed = Moles of NaOH used = 0.44 moles
Step 4: Determine the molarity of NaCl in the resulting solution
Molarity of NaCl = Moles of NaCl formed / Volume of resulting solution (in liters)
The resulting solution is obtained by adding the volumes of the HCl and NaOH solutions together:
Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution
Volume of resulting solution = 0.3 L + 0.2 L = 0.5 L
Now we can calculate the molarity of NaCl:
Molarity of NaCl = 0.44 moles / 0.5 L = 0.88 M
Therefore, the molarity of NaCl in the resulting solution is 0.88 M.