A solution is prepared by mixing 90 ml of 5 M HCl and 30 ml of 8 M HNO3. Water is then added until the final volume is 1 L. Calculate the [H+], [OH-], and the pH for this solution.
mols HCl = M x L = ?
mols HNO3 = M x L = ?
M final = total mols/total L
(H^+) = final M
pH = -log(H^+)
(H^+)(OH^-) = Kw = 1E-14
To calculate the [H+], [OH-], and pH of the solution, we need to follow these steps:
Step 1: Calculate the moles of HCl and HNO3:
Moles of HCl = Volume (in L) x Concentration (in M)
= (90 ml / 1000 ml/L) x 5 M
= 0.45 moles
Moles of HNO3 = Volume (in L) x Concentration (in M)
= (30 ml / 1000 ml/L) x 8 M
= 0.24 moles
Step 2: Calculate the total moles of H+ ions:
Total moles of H+ ions = Moles of HCl + Moles of HNO3
= 0.45 moles + 0.24 moles
= 0.69 moles
Step 3: Calculate the volume of water added:
Volume of water added = Final volume - Total volume of acid
= 1 L - (90 ml + 30 ml)
= 0.88 L
Step 4: Calculate the concentration of H+ ions:
Concentration of H+ ions = Total moles of H+ ions / Final volume
= 0.69 moles / 0.88 L
= 0.7845 M
Step 5: Calculate the concentration of OH- ions:
The solution is considered to be neutral since it contains both HCl and HNO3. Therefore, the concentration of OH- ions can be calculated using Kw (the ion product of water).
Kw = [H+][OH-] = 1 x 10^-14
[OH-] = Kw / [H+]
= 1 x 10^-14 / 0.7845
Step 6: Calculate the pH of the solution:
pH = -log[H+]
= -log(0.7845)
Using a scientific calculator, you can find the pH value.
Note: The concentration of OH- ions will be very small in this case since the solution is predominantly acidic.
To calculate the [H+], [OH-], and pH for the solution, you need to follow a series of steps:
Step 1: Calculate the amount of moles of HCl and HNO3 used.
- The molarity (M) is given for both HCl and HNO3, and the volumes are given as well (90 ml and 30 ml respectively).
- Use the formula: Moles = Molarity × Volume (in liters).
- For HCl: Moles of HCl = 5 M × (90 ml ÷ 1000 ml/L).
- For HNO3: Moles of HNO3 = 8 M × (30 ml ÷ 1000 ml/L).
Step 2: Determine the total amount of moles of H+ ions in the solution.
- Since both HCl and HNO3 dissociate in water to form H+ ions, add the moles of H+ ions from HCl and HNO3 to get the total moles.
- Moles of H+ ions = Moles of HCl + Moles of HNO3.
Step 3: Calculate the total volume of the solution.
- It is given that water is added until the final volume is 1 L.
- The total volume is the sum of the initial volumes of HCl and HNO3 solutions, plus the added water.
- Total volume = 90 ml + 30 ml + (volume of added water in liters).
Step 4: Calculate the molarity of H+ ions in the solution.
- The molarity (Molarity = Moles ÷ Volume) can be calculated using the total moles of H+ ions and the total volume of the solution.
- Molarity of H+ ions = Moles of H+ ions ÷ Total volume (in liters).
Step 5: Calculate the concentration of OH- ions.
- This can be done using Kw, which is the ion product of water at a given temperature and is equal to 1.0 × 10^-14 at 25°C.
- In neutral solutions, [H+] × [OH-] = Kw.
- Since we calculated the [H+] in step 4, we can rearrange the equation to solve for [OH-].
- [OH-] = Kw ÷ [H+] = 1.0 x 10^-14 ÷ [H+].
Step 6: Calculate the pH of the solution.
- pH is defined as the negative logarithm (base 10) of the [H+] concentration.
- pH = -log10([H+]).
Now, let's perform the calculations:
Step 1: Moles of HCl = 5 M × (90 ml ÷ 1000 ml/L) = 0.45 moles
Moles of HNO3 = 8 M × (30 ml ÷ 1000 ml/L) = 0.24 moles
Step 2: Moles of H+ ions = Moles of HCl + Moles of HNO3 = 0.45 moles + 0.24 moles = 0.69 moles
Step 3: Total volume = 90 ml + 30 ml + (volume of added water in liters) = 1 L
Since the total volume is already given as 1 L, no additional water is added.
Step 4: Molarity of H+ ions = Moles of H+ ions ÷ Total volume = 0.69 moles ÷ 1 L = 0.69 M
Step 5: [OH-] concentration = Kw ÷ [H+] = 1.0 x 10^-14 ÷ 0.69 M = 1.45 x 10^-14 M
Step 6: pH = -log10([H+]) = -log10(0.69) ≈ 0.16
Therefore, the values for this solution are:
[H+] = 0.69 M
[OH-] = 1.45 x 10^-14 M
pH = 0.16