Answer to Question #135105 in General Chemistry for Scott Ferguson

Question

I have a couple questions i need an explanation for....
1) How would we prepare 250mls of a 4M NaOH solution in water using sodium hydroxide pellets?
2) 10.75g of NaCl is dissolved in 500 mls of water to make a stock solution
- What is the concentration of Na (%w/v)?
- 25 ml of stock solution is diluted to 200 mls. concentration Na in mg/ml?
- 20 mls of resulting solution is diluted to 100 mls. concentration Na in mg/ml
- 5 mls of resulting is diluted to 25 ml with 1:1 water:nitric acid. Na of solution in mg/ml?

Answer 1

Q1. We are preparing 250mL of 4M NaOH

4M means that we are having 4mol in 1000mL of solution.

How many moles do we need then to make 250mL of the same solution?

Moles required the; $=\dfrac{Molarity x volume}{1000}= \dfrac{4molx 250mL}{1000mL} = 1mol$

But molar mass (1 mol) of NaOH is 40g

Thus you need 40g of NaOH in 250mL of solution.

You will put 40g in about 200mL of solution, stir to dissolve then transfer the contents into a 250mL volumetric flask. Add water to the mark to make it 250mL.

The solution will be 250mL of 4M NaOH

Q2. (a) You have 10.75g of NaCl in 500mL of solution

(Assuming in your question you mean concentration of Na⁺ in %w/v)

Now we need to determine mass of Na+ in the 10.75g of NaCl

Molar mass of NaCl = (23+35.5) = 58.5g/mol

Therefore it means that;

For every 58.5g of NaCl, there are 23g of Na+

What is then the mass of Na+ in 10.75g of NaCl?

Mass of Na+ $=\dfrac{10.75gx23g}{58.5g}=4.23g$

calculating concentration

w/v(%) $=\dfrac{mass of solute}{volume of solution}$ $=\dfrac{4.23g}{500mL}=8.45x10^-3g/mL$

(b) 25mL of the solution above is diluted to 200mL

In that solution,

1mL contains 8.45x10^-3g of Na

25mL will contain ? g

$=\dfrac{25mLx8.45x10^-3g}{1mL}=0.211g$ of Na

Concentration of the new solution will be 0.211g of Na in 200mL of solution

$=\dfrac{0.211g}{200mL}=1.057x10^-3g/mL$

(c) 20mL of the solution in (b) above is diluted to 100mL

1mL of the solution contains 1.057x10^-3 g of Na

20 mL would contain ? g

$=\dfrac{1.057x10^-3gx20mL}{1mL}=0.0211g$

The new concentration is therefore 0.0211g of Na in 100mL of solution

$=\dfrac{0.0211}{100mL}=2.11x10^-4g/mL$

(d) 5mL of this solution is diluted into 25mL of nitric acid solution (1:1)

Assumptions: (The new solution is 25mL and the mass of Na ions does not change)

1mL of the solution (c) above has 2.11x10^-4g of Na

5mL will have ? g

$=\dfrac{5mLx2.11x10^-4g}{1mL}=1.055x10^-3g$

This is the mass mixed with 25mL of Nitric acid. With the assumptions above

The new concentration of Na will be;

$=\dfrac{1.055x10^-3g}{25mL}=4.22x10^-5g/mL$