If a rate of 80.0 mL of an aqueous solution of potassium hydroxide concentration of 0.25 mol / L are partially neutralized by 20.0 mL of an aqueous solution of nitric acid concentration of 0.50 mol L. Determine pH of the final solution.
kOH + HNO3 -> H2O + KNO3
mols HNO3 = M x L = ?
mols KOH = M x L = ?
Look to see which has more mols then subtract the smaller from the larger. The excess will determine the pH. If acid is in excess then mols excess/total volume in L = (H^+) and convert to pH.
If KOH is in excess then excess/total volume in L = OH, convert to pOH and pH. Post your work if you get stuck
To determine the pH of the final solution, we need to calculate the concentration of the resulting ions after the reaction between potassium hydroxide (KOH) and nitric acid (HNO3). Here's how you can do it step by step:
1. Calculate the amount of moles of each reactant:
Moles of KOH = Volume (L) x Concentration (mol/L)
= 0.080 L x 0.25 mol/L
= 0.020 mol
Moles of HNO3 = Volume (L) x Concentration (mol/L)
= 0.020 L x 0.50 mol/L
= 0.010 mol
2. Determine the limiting reactant:
The limiting reactant is the one that runs out first and determines the maximum amount of product formed. In this case, both KOH and HNO3 react in a 1:1 ratio, so the limiting reactant is HNO3 because it has fewer moles (0.010 mol) compared to KOH (0.020 mol).
3. Calculate the remaining moles of the excess reactant (KOH in this case):
Remaining moles of KOH = Initial moles of KOH - Moles of HNO3
= 0.020 mol - 0.010 mol
= 0.010 mol
4. Calculate the concentration of KOH in the final solution:
Final volume = Volume of KOH + Volume of HNO3
= 0.080 L + 0.020 L
= 0.100 L
Concentration of KOH in the final solution = Remaining moles of KOH / Final volume
= 0.010 mol / 0.100 L
= 0.10 mol/L
5. Calculate the hydroxide ion concentration ([OH-]) in the final solution:
Since KOH is a strong base, it dissociates completely in water to produce one hydroxide ion for every molecule of KOH.
[OH-] = Concentration of KOH
= 0.10 mol/L
6. Calculate the pOH of the final solution:
pOH = -log10([OH-])
= -log10(0.10)
= 1
7. Calculate the pH of the final solution:
pH + pOH = 14 (at 25°C)
pH = 14 - 1
= 13
Therefore, the pH of the final solution is 13.