Calculate the pH when the following volumes of 0.20 M LiOH are added into 50 mL of a 0.1 M HCl solution. Both LiOH and HCl are strong electrolytes.
a) 0 mL
b) 10 mL
c) 24.6 mL
d) 25 mL
e) 25.6 mL
f) 30 mL
g) 40 mL
This is a strong acid/base titration. Balance your equation for the reaction and calculate if you have excess H30+ or OH-
Remember that pH=-log[H30+]
and pOH = -log[OH-]
pOH = 14 - pH
Where [H30+] and [OH-] is the concentration of each substance in moles/liter
To calculate the pH when different volumes of LiOH are added to the HCl solution, we need to consider the neutralization reaction between the two chemicals. The reaction between LiOH and HCl can be represented as:
LiOH + HCl -> LiCl + H2O
In this reaction, LiOH reacts with HCl to produce LiCl (lithium chloride) and water (H2O).
Before any LiOH is added, we have a 50 mL solution of 0.1 M HCl. Since HCl is a strong acid, it completely dissociates into its ions, H+ and Cl-. Therefore, the concentration of H+ ions in the solution is 0.1 M.
Now, let's calculate the pH for each case:
a) 0 mL:
When no LiOH is added, the solution remains as a 50 mL 0.1 M HCl solution. The concentration of H+ ions remains the same at 0.1 M. To calculate the pH, we can use the formula: pH = -log[H+].
pH = -log(0.1) = 1.
b) 10 mL:
If 10 mL of 0.20 M LiOH is added, we can determine the number of moles of LiOH added and the resulting concentrations of Li+ and OH- ions.
Moles of LiOH added = volume (L) x concentration (M)
Moles of LiOH added = 0.01 L x 0.20 M = 0.002 mol
Since LiOH is a strong base, it completely dissociates into Li+ and OH- ions. Therefore, the concentration of Li+ ions and OH- ions in the solution will be equal to the concentration of LiOH added.
Concentration of Li+ and OH- ions = moles / volume
Concentration of Li+ and OH- ions = 0.002 mol / (0.01 L + 0.05 L) = 0.002 mol / 0.06 L = 0.0333 M
Since LiOH is a strong base, it reacts with H+ ions to produce water (neutralization reaction). Since the concentration of LiOH added is greater than that of HCl, all HCl will be neutralized.
The number of moles of HCl neutralized = moles of LiOH added = 0.002 mol
The number of moles of H+ remaining = initial moles of H+ - moles of HCl neutralized
The number of moles of H+ remaining = initial concentration x volume - 0.002 mol
The number of moles of H+ remaining = 0.1 M x 0.05 L - 0.002 mol = 0.003 mol
The final concentration of H+ ions = moles / volume
The final concentration of H+ ions = 0.003 mol / (0.01 L + 0.05 L) = 0.003 mol / 0.06 L = 0.05 M
To calculate the pH, we can use the formula: pH = -log[H+].
pH = -log(0.05) = 1.3.
You can follow the same procedure to calculate the pH for cases c), d), e), f), and g). Just substitute the volumes of LiOH added and repeat the steps outlined above.