Calculate the pH of the solution formed when 20mL of 0.100 mol/L HNO_3 is added to 30 mL of 0.050 mol/L KOH.
millimols HNO3= mL x M = 20 x 0.1 = 2.0
millimols KOH = 30 x 0.05 = 1.0
Total volume is 20 mL + 30 mL = 50 mL
.............................KOH + HNO3 ==> KNO3 + H2O
Initial......................1.0..........0...............0...........0
add......................................2.0....................................
Change.................-1.0........-1.0...........1.0............1.0
Equilibrium...............0...........1.0............1.0
So the chart above tells you that when the reaction is complete you have 1.0 millimols left of HNO3 in 50 mL. Since millimols = mL x M then M = millimols/mL = 1.0 mmols/50 mL = 0.02 M
Use pH = -log(HNO3) to convert to pH. Post your work if you get stuck.
To calculate the pH of the solution formed when an acid and a base are mixed, we can use the concept of neutralization. In this case, nitric acid (HNO3) and potassium hydroxide (KOH) will react to form water (H2O) and potassium nitrate (KNO3).
The balanced chemical equation for this reaction is:
HNO3 + KOH -> H2O + KNO3
To calculate the pH of the resulting solution, we need to consider the stoichiometry of the reaction and the concentrations of the acid and base.
Step 1: Determine the limiting reagent
To find the limiting reagent, we compare the moles of each reactant using the balanced equation. The reactant that gives the fewer moles of product is the limiting reagent.
In this case, the balanced equation tells us that 1 mole of HNO3 reacts with 1 mole of KOH. So, the moles of HNO3 and KOH are the same as the volumes given in the question.
20 mL of HNO3 (0.020 L) x 0.100 mol/L = 0.002 moles of HNO3
30 mL of KOH (0.030 L) x 0.050 mol/L = 0.0015 moles of KOH
Since HNO3 has a greater number of moles, it is the limiting reagent.
Step 2: Calculate the moles of HNO3 reacted
Since HNO3 is the limiting reagent, all of the HNO3 will react. Therefore, the moles of HNO3 reacted are equal to the initial moles of HNO3.
Moles of HNO3 reacted = 0.002 moles
Step 3: Calculate the moles of HNO3 remaining
To calculate the moles of HNO3 remaining, we subtract the moles of HNO3 reacted from the initial moles of HNO3.
Moles of HNO3 remaining = Initial moles of HNO3 - Moles of HNO3 reacted
= 0.002 moles - 0.002 moles
= 0 moles
Step 4: Calculate the moles of OH- reacted
Since KOH is in excess, all of the HNO3 that reacts will be neutralized by the KOH in a 1:1 ratio. Therefore, the moles of OH- reacted are equal to the moles of HNO3 reacted.
Moles of OH- reacted = Moles of HNO3 reacted
= 0.002 moles
Step 5: Calculate the moles of OH- remaining
To calculate the moles of OH- remaining, we subtract the moles of OH- reacted from the initial moles of KOH.
Moles of OH- remaining = Initial moles of KOH - Moles of OH- reacted
= 0.0015 moles - 0.002 moles
= -0.0005 moles
Since the result is negative, it means that there are no moles of OH- remaining. The excess HNO3 has fully neutralized the KOH.
Step 6: Calculate the net ionic equation
Since the reaction goes to completion, the H+ from HNO3 reacts with the OH- from KOH to form H2O. Therefore, the net ionic equation is:
H+ + OH- -> H2O
Step 7: Calculate the concentration of H+
Since 0.002 moles of HNO3 were initially present in a total volume of 50 mL (20 mL + 30 mL), we can calculate the concentration of H+ using the following formula:
[H+] = Moles of H+ / Total volume of solution
= 0.002 moles / 0.050 L
= 0.04 mol/L
Step 8: Calculate the pH
The pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the concentration of H+.
pH = -log[H+]
= -log(0.04)
≈ 1.4
Therefore, the pH of the solution formed when 20 mL of 0.100 mol/L HNO3 is added to 30 mL of 0.050 mol/L KOH is approximately 1.4.