If 400 mL of 3.2 molar HCl solution is added to 300 mL of 2 molar NaOH solution, what will be the molarity of NaCl in the resulting solution? Answer in units of M
HCl + NaOH ==> NaCl + H2O
mols HCl = M x L = 3.2 x 0.4 = 1.28
mols NaOH = M x L = 2 x 0.3 = 0.6
mols NaCl formed = 0.6
M NaCl = mols/L = 0.6/(0.4+0.3) = ?
To find the molarity of NaCl in the resulting solution, we can use the concept of stoichiometry.
First, let's calculate the number of moles of HCl and NaOH in their respective solutions.
For the HCl solution:
Molarity of HCl = 3.2 M
Volume of HCl solution = 400 mL = 0.4 L
Using the formula C = n/V, where C is the molarity, n is the number of moles, and V is the volume in liters, we can rearrange the formula to solve for n:
n (moles) = C (molarity) x V (volume)
Number of moles of HCl = 3.2 M x 0.4 L = 1.28 moles
Similarly, for the NaOH solution:
Molarity of NaOH = 2 M
Volume of NaOH solution = 300 mL = 0.3 L
Number of moles of NaOH = 2 M x 0.3 L = 0.6 moles
Now, let's consider the balanced chemical equation for the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that the mole ratio between HCl and NaCl is 1:1. This means that for every 1 mole of HCl reacted, we will get 1 mole of NaCl.
Since we have 1.28 moles of HCl and assuming all of it reacts completely, we will have 1.28 moles of NaCl.
To find the molarity of NaCl in the resulting solution, we need to calculate the total volume of the resulting solution. By adding the volumes of the HCl and NaOH solutions, we get:
Total volume = 400 mL + 300 mL = 700 mL = 0.7 L
Now, we can calculate the molarity of NaCl:
Molarity of NaCl = number of moles of NaCl / total volume
Molarity of NaCl = 1.28 moles / 0.7 L = 1.83 M
Therefore, the molarity of NaCl in the resulting solution is 1.83 M.
HCL + NAOH -> NaCL + H2)
Mols HCL = M x L = 3.2 x 0.4 = 1.28
Mols NaOH = M x L = 2 x 0.3 = 0.6
mols NaCl formed = 1.28
M NaCl = mols/L = 1.28/(.4 + .3) = 1.829