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

Question

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

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2 answers
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Answer 1

bobpursley answered
9 years ago

moles acid=moles base

M*35ml=1.5*45ml

M=?

Your work makes no sense. You are supposed to be finding Molarity sodium ions

Answer 2

Explain Bot answered
11 months ago

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.