Lye concentration
Measurements
| Amount | Breakdown |
|---|---|
| Low water \(50\%\space LC\) | \(50\%\space NaOH\) (Lye) |
| \(50\%\space H_20\) (Water) | |
| Medium water \(33\%\space LC\) | \(33\%\space NaOH\) (Lye) |
| \(67\%\space H_20\) (Water) | |
| High water \(25\%\space LC\) | \(25\%\space NaOH\) (Lye) |
| \(75\%\space H_20\) (Water) |
"Lye" A.K.A. "Caustic soda", is the chemical Sodium hydroxide (\(NaOH\))
Calculations
Page 1
\[
C=\frac{n}{V},\space\space n=\frac{N}{N_A}=\frac{m}{M}
\]
If \(m\) is in \([g]\), then \(M\) in \([g * mol^{-1}]\) and so on - must cancel? to give \(mol\) without magnitude change.
m [g]
M [g*mol_1]
\[
\therefore C=\frac{n}{V},\space\space n=\frac{n}{M}
\]
so then:
\[
C=\frac{m}{VM}
\]
\[
M_{NaOH} = 22.99 + 16 + 1.008 = 39.998\space\space g \cdot mol^{-1}
\]
\[
\therefore M_{NaOH}=40\space\space g \cdot mol^{-1}
\]
Page 2
\[
n_L\space |\space const
\]
Let the following variables have the values:
\[
L\congruent NaOH,\space\space W\congruent H_20 \space\space V=V_w
\]
\[
C_L = \frac{n_L}{V} = \frac{m_L}{V_{w}M_{L}}
\]
\[
\therefore n_L = \frac{m_L}{M_L}
\]
\[
P_w = \frac{m_w}{V}
\]
where \(P_w\) is for water density (by mass) at \(20^{^O}C \congruent 293.15K = temperature\) and \(1013.25\space hPa = pressure\) . (see 1)
m [kg*m^-3]
M [g*L^-1]
\[
\therefore P_w = 998.23 kg \cdot m^{-3}
\]
\[
\arrow P_w = 998.23 g \cdot L^{-1}
\]
\[
1kg \congruent 1000g,\space\space 1m^{3} \congruent 1 * (10^{1})^{3}\space dm^{3} = 1000L
\]
\[
\therefore \frac{1000}{1000} = 1,\space 1kg \cdot m^{-3} = 1\space g \cdot L^{-1}
\]
Therefore the mass density of the water is:
\[
P_w = \therefore 998.23\space g \cdot L^{-1}
\]
- According to van.physics.illinois.edu
Page 3
\[
M_{H_2O} = 2 * 1.008 + 16
\]
\[
M_{H_2O} = 18.016\space g \cdot mol^{-1}
\]
\[
V = m_L * \frac{(1-\frac{P_L}{100})M_w}{(\frac{P_L}{100})P_wM_L}
\]
where \(L\) is for \(NaOH\) and \(w\) is for \(H_2O\) (water).
For Litres [L]
\[
V = m_L * \frac{100-P_L}{P_L} * (\frac{18.016}{998.23*40})
\]
For Litres [mL]
\[
V = m_L * \frac{100-P_L}{P_L} * (\frac{18.016*10^3}{998.23*40})
\]