For a solution of known volume and concentration (molarity in this case), the volume needed to dilute the solution to a desired concentration may be found using the formula:

\(\displaystyle M_{1}V_{1} = M_{2} V_2\)

Where \(\displaystyle M_1\) and \(\displaystyle M_2\) are the initial and final concentrations, and \(\displaystyle V_1\) and \(\displaystyle V_2\) are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

\(\displaystyle 0.050 M * 0.750 L = 0.015 M * V_2\)

Solving for \(\displaystyle V_2\):

\(\displaystyle V_2 = \frac{0.050 M * 0.750L}{0.015 M} = 2.5 L\)

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume (\(\displaystyle V_1\)) from the final volume (\(\displaystyle V_2\)):

\(\displaystyle V_2 - V_1 = 2.5 L - 0.750 L = 1.75 L\)

1.75L of water must be added to 750mL of 0.050M \(\displaystyle NaCl\) in order to achieve a final concentration of 0.015M

Since \(\displaystyle HCl\) is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

\(\displaystyle HCl\rightarrow H^{+} + Cl^{-}\)

\(\displaystyle H^+\) and \(\displaystyle Cl^-\) are produced in a 1:1 ratio to total dissolved \(\displaystyle HCl\), so the concentration of \(\displaystyle HCl\) in solution is the same as the concentration of \(\displaystyle H^+\):

\(\displaystyle \left [ HCl\right ] = \left [ H^{+}\right ] = \left [ Cl^{-}\right ]\)

pH is related to the concentration of \(\displaystyle H^+\):

\(\displaystyle pH = -log[H^{+}]\)

\(\displaystyle pH = -log[0.025] = 1.60\)

Osmotic pressure is represented by:

\(\displaystyle \small \Pi = iMRT\)

Where \(\displaystyle \small i=\) Van’t hoff factor, \(\displaystyle \small M = Molarity\), \(\displaystyle \small R =\) gas constant \(\displaystyle \left(0.08206 \frac{L\cdot atm}{mol\cdot K}\right)\), \(\displaystyle \small T =\) temperature in \(\displaystyle \small K\). The Van’t hoff factor is a unitless number that represents the amount of ionic species that the compound \(\displaystyle \small (C_{2}H_{6})\) will dissociate in solution. \(\displaystyle \small C_{2}H_{6}\) is part of a large group of molecules classified as hydrocarbons which normally do not dissociate at all in solution. Therefore, \(\displaystyle \small i = 1\).

Plug in known values and solve.

\(\displaystyle \Pi = 1\cdot 5\cdot 0.08206\frac{L\cdot atm}{mol\cdot K}\cdot 283\)

\(\displaystyle \Pi= 116atm\)

In order to calculate the concentration, we must use molarity formula:

\(\displaystyle Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}\)

We must use the molecular weight of \(\displaystyle NaOH\) to calculate the moles of solute:

\(\displaystyle MW_{NaOH}=(23\frac{g}{mole})+(16\frac{g}{mole})+(1\frac{g}{mole})=40\frac{g}{mole}\)

\(\displaystyle 22g*\frac{mole}{40g}=0.55\ moles\ NaOH\)

\(\displaystyle \frac{0.55\ moles\ NaOH}{0.110L}=5M\)

In order to calculate the concentration, we must use molarity formula:

\(\displaystyle Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}\)

We must use the molecular weight of \(\displaystyle CH_{3}CH_{2}OH\) to calculate the moles of solute:

\(\displaystyle MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ O)\)

\(\displaystyle MW_{CH_{3}CH_{2}OH}=2(12\frac{g}{mole})+6(1g\frac{g}{mole})+(16\frac{g}{mole})=46 \frac{g}{mole}\)

\(\displaystyle 40.0g*\frac{mole}{46g}=0.87\ moles\ CH_{3}CH_{2}OH\)

\(\displaystyle \frac{0.87\ moles\ CH_{3}CH_{2}OH}{0.050\L}=17M\)

In order to calculate the number of milliliters, we must first determine the number of moles in 5 grams of \(\displaystyle NaCl\) using its molecular weight as a conversion factor:

\(\displaystyle MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)\)

\(\displaystyle MW_{NaCl}=(23\frac{g}{mole})+(35\frac{g}{mole})=58\frac{g}{mole}\)

\(\displaystyle 5g*\frac{mole}{58g}= 0.086\ moles\)

Using the concentration units as a conversion factor and the number of moles calculated, the number of milliliters can be calculated:

\(\displaystyle 0.086 moles * \frac{L}{10\ moles}= 0.0086\ L = 8.6\ mL\)

Use the dilution formula:

\(\displaystyle M_{1}V_{1}=M_{2}V_{2}\)

Rearranging this equation gives:

\(\displaystyle \frac{M_{1}V_{1}}{V_{2}}=M_{2}\)

Plugging in the values gives:

\(\displaystyle M_{2}=\frac{0.500M * 10mL}{20mL}= 0.25M\)

Therefore, after diluting the solution to 20mL, the solution concentration would be lowered from 0.50M to 0.25M.

Solutes that dissociate completely in a solution are called strong electrolytes. Weak electrolytes stay paired to some extent in solutions. As a result, strong electrolytes include ionic compounds and strong acid and bases.

Solubility product, \(\displaystyle K_{sp}\), is the product of ion concentrations at equilibrium in a saturated solution of salt. All other definitions are true.

Balancing equations:

there is 1 NaOH and 1 H₂SO₄ in the reactants; and 1 H₂O and 1 Na₂SO₄ in the products

steps:

1) balance the Na first; get 2 NaOH

2) balance H next; get 1 H₂SO4, 2 H₂O

3) check to make sure that O and S balance