What is the concentration of Na+ ions if 35.0 mL of 2.50 M NaOH is reacted with 47.0 mL of 1.50 M
HCl?
I did M=Moles/Volume
Moles=M*V
Moles=2.50M*0.035= 0.0875
Am I right? If I am right, I don't know what to do next. Please help
moles acid=moles base
M*35ml=1.5*45ml
M=?
Your work makes no sense. You are supposed to be finding Molarity sodium ions
Yes, you are correct in calculating the number of moles. To find the concentration of Na+ ions, you need to consider the reaction between NaOH and HCl.
The balanced equation for the reaction is:
NaOH + HCl → NaCl + H2O
From the balanced equation, you can see that for every 1 mole of NaOH, 1 mole of Na+ ions is produced.
Since you have calculated that the number of moles of NaOH is 0.0875, this means that you have 0.0875 moles of Na+ ions.
Now, let's find the total volume of the solution after the reaction has occurred. The total volume can be calculated by adding the volumes of NaOH and HCl:
Total volume = 35.0 mL + 47.0 mL = 82.0 mL
Next, you need to convert the total volume to liters:
Total volume = 82.0 mL = 82.0 mL * (1 L / 1000 mL) = 0.0820 L
Finally, you can calculate the concentration (C) of Na+ ions using the formula:
C = moles / volume
C = 0.0875 moles / 0.0820 L = 1.07 M
Therefore, the concentration of Na+ ions in the resulting solution is 1.07 M.