Unit 4 - Solutions and Solubility

36.525### Factors effecting the rate of dissolving:

• Temperature
  • Hot is faster
• Agitation
• Surface area

Strongly Polar molecules

Acids

Slightly Polar molecules

Molecules only dissolve but DO NOT form ions
e.g. Glucose, ethanol

Non-Polar molecules

Molecules will not dissolve in water “like dissolves like”
e.g. $CC{l}_4$ will not dissolve in water but will in $C{S}_2$

The dissolving of Molecular compounds

Miscible

Readily Able to dissolve

Immiscible

Does not dissolve

Surfactants

• Encourages oil and water to mix
• Has a negative charged “water-loving” head and a “grease-loving” hydrocarbon tail

Percentage Concentration

• Percentage volume by volume (% v/v)
• Percentage mass by volume (% m/v)
  • solute mass is expressed in grams
  • solution is expressed in mL
• Percentage mass by mass (% m/m)

Very Low Concentraations

Use PPM ($1:10^6$) or PPB ($1:10^9$) or PPT ($1:10^{12}$)

• Mass of solution is assumed to be the same as the density of the solvent
  (note 1 mL water → 1 GRAM not mg )

$$C_{ppm}=\frac{m_{solute}}{m_{solution}}\cdot 10^{6}ppm$$ $$C_1\ast V1=C_2\ast ,\ast V_2-\text{only works during dilution, not stoich}$$

TODO

• Review dipoles
• Revire 379
• Figure out Silver nitrate problem

e.g.

$$\begin{array}{r}x\ast 18=2\ast 0.1\\ \\ \frac{2\ast 0.1}{18}=x\\ x=0.01111111111\\ x=1.1\cdot 10^{-}2\\ \end{array}$$

$\therefore$ the amount of concentrated acid needed would be 1.1E-2 L

2

$$\begin{array}{rl}& 3.25\cdot x=1.30\cdot 250mL\\ & \\ & 3.25\cdot x=1.30\cdot \frac{250}{1000}L\\ & \\ & x=\frac{1.30\cdot 0.25L}{3.25}\\ & x_L=0.1L\\ & x_{mL}=0.1\cdot 1000\\ & x=100mL\end{array}$$

Other questions

$$\begin{array}{rl}& 0.1L\cdot 19.8g\\ & \frac{19.8}{74.5513}=0.26558893mo{l}_{KCl}\\ & \frac{molC{l}^{-}}{molKCl}=1:1\text{mole ratio}\\ & 0.26558893mo{l}_{KCl}\cdot 1=0.26558893mo{l}_{Cl}\\ & \frac{0.26558893}{0.1L}=2.6558893mol/L\end{array}$$

$\therefore$ the concentration of chloride ions in this problem is 2.66mol/L.
twooo

$$\begin{array}{r}\\ CaC{l}_2\to C{a}^{+}+C{l}_2^{-}\\ \frac{26.5}{110.98}=0.23878176248mol\\ \frac{2molC{l}_2}{1molCaC{l}_2}=2:1\text{ratio}\\ \\ 0.23878176248\ast \frac{2}{1}=0.477563525mol\\ c\frac{0.477563525}{0.15}\\ 3.18375683333\frac{mol}{L}\end{array}$$

NEW

$$\begin{array}{rl}& MgC{L}_{2(aq)}+2AgN{o}_{3(aq)}\to 2AgC{l}_{(s)}+Mg(N{O}_3{)}_{2(aq)}\\ & c=0.25mol/L=0.30mol/L\\ & 0.3\ast 0.060=0.018mo{l}_{AgN{o}_3}\end{array}$$

Ask dore about diff between

0.1 / 18 = 0.00555555556
2 * 0.00555555556 = 0.0111111111

yet the other way gave me a flat 0.1