45.00 mL of 1.55×10^−2 M HI(aq) is mixed with 85.00 mL of 1.08×10^−2 M KOH(aq). What is the pH of the solution?
It depends upon which material (acid or base) is in excess.
So calculate moles HI. M x L = moles.
Calculate moles KOH. M x L = moles.
Since HI is a strong acid and KOH is a strong base, and the reaction is 1 mole to 1 mole, subtract the smaller number of moles from the larger number of moles. That will tell you which is in excess, then divide the answer by the number of L of solution (you will need to add volume of HI to volume of KOH to obtain final volume.)
i tried the -log of the solution
which is -log(0.001696153) = 2.77 pH but that is incorrect!
Thats cause you found POH, subtract that from 14
yes 11.23 is correct
Well, looks like we have a chemistry question here! Mixing an acid and a base, huh? Let's dive in and find the pH of this solution!
First, let's calculate the moles of HI and KOH in the given volumes. To find the moles, we can use the formula: moles = concentration (M) × volume (L).
So, for HI, we have:
moles of HI = (1.55×10^−2 M) × (0.04500 L) = 6.975×10^−4 mol
For KOH, we have:
moles of KOH = (1.08×10^−2 M) × (0.08500 L) = 9.180×10^−4 mol
Now it's time to determine which species will react to form water. By looking at the balanced chemical equation for the reaction between HI and KOH, we see that they react in a 1:1 ratio. This means that all the HI will be used up, leaving us with an excess of KOH.
The reaction between HI and KOH forms water (H2O) and a salt (KI). Since we have excess KOH, we can assume all HI is consumed.
Since HI is a strong acid and KOH is a strong base, we can consider the resulting solution to be a salt, which in this case is KI.
Now, to find the pH, we need to calculate the concentration of OH- ions from the excess KOH. Since the KOH is strong, it completely dissociates in water, so we can use the concentration of KOH to determine the concentration of OH- ions.
The concentration of OH- ions in the solution is given by:
OH- concentration = moles of KOH / total volume of the solution
The total volume of the solution is the sum of the given volumes:
total volume = 45.00 mL + 85.00 mL = 130.00 mL = 0.13 L
Thus, OH- concentration = (9.180×10^−4 mol) / (0.13 L) ≈ 7.062×10^−3 M
Now we have the concentration of OH- ions. To determine the pOH, we can use the formula: pOH = -log10(OH- concentration).
Using that formula, we have:
pOH ≈ -log10(7.062×10^−3) ≈ 2.15
Finally, to find the pH, we can use the relationship pH + pOH = 14. So, rearranging for pH, we have: pH = 14 - pOH.
Substituting the value of pOH, we get:
pH = 14 - 2.15 ≈ 11.85
Therefore, the pH of the solution is approximately 11.85.
To find the pH of the solution, we first need to determine the concentration of H+ ions in the solution. We can do this by calculating the moles of H+ ions from the reaction between HI and KOH and then dividing it by the total volume of the solution.
The balanced equation for the reaction between HI and KOH is:
HI(aq) + KOH(aq) → H2O(l) + KI(aq)
From the equation, we can see that one mole of HI reacts with one mole of KOH to produce one mole of water and one mole of KI.
First, calculate the number of moles of HI:
moles of HI = volume of HI (in L) × concentration of HI (in mol/L)
= 0.04500 L × 1.55×10^−2 mol/L
= 6.975×10^−4 mol
Next, calculate the number of moles of KOH:
moles of KOH = volume of KOH (in L) × concentration of KOH (in mol/L)
= 0.08500 L × 1.08×10^−2 mol/L
= 9.18×10^−4 mol
Since HI and KOH react in a 1:1 ratio, the moles of H+ ions present in the solution will be the same as the moles of HI used. Therefore, the moles of H+ ions in the solution are 6.975×10^−4 mol.
Now, we need to calculate the total volume of the solution:
total volume of solution = volume of HI + volume of KOH
= 0.04500 L + 0.08500 L
= 0.1300 L
Finally, we can calculate the concentration of H+ ions in the solution:
concentration of H+ ions = moles of H+ ions / total volume of solution
= 6.975×10^−4 mol / 0.1300 L
= 5.35×10^−3 M
The concentration of H+ ions in the solution is 5.35×10^−3 M.
To find the pH of the solution, we use the pH formula:
pH = -log[H+]
Substituting the concentration of H+ ions into the equation, we have:
pH = -log(5.35×10^−3)
= -(-2.27)
= 2.27
Therefore, the pH of the solution is 2.27.