what is the pH of a solution prepared by adding 0.65mol of LiOH(aq) to 1.0L of 0.30mol/L HNO3(aq)>
To determine the pH of a solution prepared by mixing LiOH(aq) and HNO3(aq), you need to know the molar concentrations of both LiOH and HNO3, as well as their reaction equation.
The balanced chemical equation for the reaction between LiOH and HNO3 is as follows:
LiOH + HNO3 -> LiNO3 + H2O
From the equation, you can see that 1 mole of LiOH reacts with 1 mole of HNO3 to produce 1 mole of LiNO3 and 1 mole of water.
Given:
Molar concentration of HNO3 (C1) = 0.30 mol/L
Volume of HNO3 (V1) = 1.0 L
Moles of LiOH (n2) = 0.65 mol
First, calculate the moles of HNO3 (n1) using the following equation:
n1 = C1 x V1
n1 = 0.30 mol/L x 1.0 L
n1 = 0.30 mol
Since 1 mole of LiOH reacts with 1 mole of HNO3, and both are in 1:1 ratio according to the equation, the moles of HNO3 in excess can be determined as:
n_excess = n1 - n2
n_excess = 0.30 mol - 0.65 mol
n_excess = -0.35 mol (negative value indicating excess LiOH)
Since LiOH is a strong base and reacts completely with HNO3, the remaining LiOH is in excess and won't contribute to the pH directly.
To calculate the concentration of the H+ ions (also known as the [H+]), you need to consider only the HNO3 reacted. The equation shows that 1 mole of HNO3 produces 1 mole of H+ ions:
[H+] = n1 / V1
[H+] = 0.30 mol / 1.0 L
[H+] = 0.30 mol/L
Now, to calculate the pOH, you can use the formula:
pOH = -log10([OH-])
Since LiOH is a strong base, it reacts completely to produce OH- ions. In this case, since LiOH is in excess, the OH- concentration is equal to the concentration of the initially added LiOH.
[OH-] = n2 / V1
[OH-] = 0.65 mol / 1.0 L
[OH-] = 0.65 mol/L
Now, calculate the pOH:
pOH = -log10(0.65)
pOH ≈ 0.19
Finally, calculate the pH using the relation:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.19
pH ≈ 13.81
So, the pH of the solution is approximately 13.81.