what is the pH of a solution prepared by adding 0.50 mol KOH to 1.0 L of 0.30 M HNO3?
Since:
mol of HNO3 = 0.30 x 1 = 0.3 mol
And mol of KOH bigger than mol of HNO3
Then:
The solution has more OH therefore basic
POH = - log [OH]
[OH] = [OH] - [H]
POH = - log [0.2]
POH = 0.7
PH + POH = 14
PH = 14 - 0.7 = 13.3
How is there no answer to this...SOMEONE HELP PLEASE
its 13.30 and i only know that bc i saw the answer key
Well, if you're looking for a pH that will make you laugh, this equation might just do the trick! Let's find out.
First, we need to calculate the concentration of H+ ions produced by the reaction between KOH and HNO3. Since we have 0.50 mol of KOH and 1.0 L of 0.30 M HNO3, the number of moles of H+ ions produced will be equal to the number of moles of KOH added.
In this case, we have 0.50 mol of H+ ions. Now, let's calculate the molarity of H+ ions in the solution by dividing the number of moles by the volume:
Molarity (M) = moles/volume
Molarity (H+) = 0.50 mol / 1.0 L
Molarity (H+) = 0.50 M
Now, the pH can be calculated using the equation:
pH = -log10(Molarity of H+)
pH = -log10(0.50) ≈ 0.30
So, the pH of this solution is approximately 0.30, but remember, my friend, laughter is always the best solution!
To find the pH of the solution, we need to determine the concentration of the resulting solution after the reaction between KOH and HNO3.
First, let's write down the balanced equation for the reaction between KOH and HNO3:
KOH + HNO3 -> KNO3 + H2O
In this reaction, 1 mole of KOH reacts with 1 mole of HNO3, producing 1 mole of KNO3 and 1 mole of water.
Given that 0.50 moles of KOH is added to 1.0 L of 0.30 M HNO3, we need to determine the concentration of the resulting solution.
Step 1: Calculate the moles of HNO3 in the initial solution.
moles HNO3 = (0.30 M) x (1.0 L) = 0.30 moles
Step 2: Determine the limiting reagent.
Since the stoichiometry of the reaction is 1:1 between KOH and HNO3, the limiting reagent will be HNO3 since we have 0.30 moles of it and only 0.50 moles of KOH. This means that all 0.30 moles of HNO3 will react with 0.30 moles of KOH, leaving 0.20 moles of KOH unreacted.
Step 3: Calculate the moles of water formed.
Since the stoichiometry of the reaction is 1:1 between HNO3 and water, the moles of water formed is also 0.30 moles.
Step 4: Calculate the concentration of the resulting solution.
The total volume of the resulting solution is still 1.0 L since we didn't add any additional volume. The moles of water formed (0.30 moles) will remain dissolved in the solution, resulting in a final concentration of 0.30 M. Therefore, the pH of the resulting solution will be determined by the concentration of water.
The concentration of water can be approximated using the autoionization of water, which is Kw = [H+][OH-] = 1.0x10^-14 at 25°C.
Since water is neutral, the concentration of [H+] will be equal to the concentration of [OH-]. Let's denote it as x.
So, x^2 = 1.0x10^-14
Taking the square root of both sides, we get: x = 1.0x10^-7
The pH of a solution is given by the equation pH = -log[H+].
Substituting the value of [H+], we get: pH = -log(1.0x10^-7) = 7.
Therefore, the pH of the solution prepared by adding 0.50 mol KOH to 1.0 L of 0.30 M HNO3 is 7.