What is the concentration of this NaOH solution?

Question

You measure 2.95 mL of a 50% NaOH solution by weight (its density is 1.53 g mL^-1) and dilute it to 500 mL total volume. What is the concentration of this NaOH solution? Write your answer to four decimal places (X.XXXX).

Answer 1

I would start by finding the concentration of the original NaOH solution in molarity (mol/L or mmol/L or mol/mL). You can do this by using the density of the solution to find the mass (in grams) of NaOH present in the original solution. Once you have found the grams of NaOH in the original solution, divide by the mL of the solution (2.95 mL). Now, you have the molarity of the original solution (in mM if you divided by mL). Also note that density is mass/ volume.

Next, use the formula C1V1=C2V2 where C1= the concentration of the original solution, V1 = the initial volume, and C2 and V2 correspond to the solution after dilution. This allows you to solve for the final concentration in mM or M depending on the units of your original molarity.

Hope this helps!

Answer 2

Concentration can be expressed in several ways. Often, molarity (M) is used. M = mol solute/L solution.

2.95 mL solution X 1.53 g/mL X 0.50 g NaOH/g solution = 2.25675 g NaOH

Molar mass NaOH = 22.990 + 15.999 + 1.008 = 39.997 g/mol

mol NaOH = 2.25675 g/(39.997 g/mol) = 0.056423

L NaOH solution = 500 mL/(1000 mL/L) = 0.500 L

Therefore, the molarity is 0.056423/0.500 = 0.112846 M, and rounded to 4 decimal places = 0.1128 M NaOH

I would state that, based on the given data, 4 decimal places is not justified according to rules for sig figs.

None of the given data (excluding molar mass) has 4 sig figs--at most 3 from 2.95 and 1.53. Technically, 50 and 500 have only 1 sig fig each; they would need to be specified as 50.00 or 500.0 in order to have 4 sig figs.

Other units of concentration could be wt/wt %, wt/vol %, or molality, and would require separate calculations.