HNO3 with pH 3 is mixed with 2 L KOH with pH 12. assuming volumes are additive what is the pH of the final solution?
I have absolutely no idea how to do this. I do know the answer, though. I have a test tomorrow and I would really appreciate it if someone could help!!!!
Why didn't you share the answer? Then we would know if I did it right or not?
However, I don't think you copied all of the problem. HOW MUCH HNO3 with a pH of 3. It makes a huge difference.
Here is how you do it.
mL HNO3 x M = millimoles HNO3. (pH 3 makes M = 1 x 10^-3) = millimoles H^+..
mL KOH x M = millimoles KOH = 2000 x 1 x 10^-2 (pH = 12, pOH = 2; therefore, OH^- = 1 x 10^-2 M).
So which millimoles is the greater, the H^+ or the OH^-.
Subtract the smaller from the larger divide the difference by TOTAL mL and that will be either the H^+ or the OH^- and you can get pH or pOH from that.
oh, sorry!
its 2 liters HNO3 and the answer is 11.65
so it would be:
-log [(20 - 2e-8) / 4000] = 2.3 = [OH]
14 - 2.3 = 11.65 (without the rounding of the 2.3)
why, though, would you subtract it from the other?
HNO3 + KOH ==> KNO3 + H2O
moles HNO3 = 2000 x 1 x 10^-3 = 2 mmoles.
moles KOH = 2ooo x 1 x 10^-2 = 20 mmoles.
Why subtract. This is an acid/base reaction, they neutralize each other. ALL of the HNO3 is GONE and you are left with an excess of 20-2 =18 mmoles KOH.
So (OH^-) = 18 mmoles/4000 mL = 0.00450 M.
pOH = 2.35 and pH = 11.65. :-).
Isn't chemistry FUN?!
Sure, I can help you with that! To find the pH of the final solution, we need to apply the principles of acid-base neutralization. The HNO3 (nitric acid) and KOH (potassium hydroxide) will react and form water (H2O) and a salt (KNO3). In this case, since both the acid and the base are strong, we can assume complete ionization of both.
First, let's calculate the concentration of H+ and OH- ions in each solution using the pH values given. The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions. The relationship between pH and H+ concentration is expressed by the formula: pH = -log[H+].
For the HNO3 solution with pH 3, we know that the concentration of H+ ions is 10^(-pH) = 10^(-3) = 0.001 M.
Similarly, for the KOH solution with pH 12, the concentration of OH- ions is 10^(-pOH) = 10^(-14 +14-12) = 10^(-2) = 0.01 M. (Note: pOH = 14 - pH).
Now, let's determine the moles of H+ and OH- ions in each solution. Since we know the volume of the KOH solution is 2 L, we can directly calculate the number of moles of OH- ions:
Moles of OH- ions = concentration of OH- ions × volume of solution
= 0.01 M × 2 L
= 0.02 moles
The HNO3 solution has an unknown volume, so we'll represent it as "V" in liters. The number of moles of H+ ions is then:
Moles of H+ ions = concentration of H+ ions × volume of solution
= 0.001 M × V L
= 0.001V moles
According to the principle of neutralization, the moles of OH- ions will react with an equal number of moles of H+ ions. Therefore, we can equate the two expressions and solve for "V":
0.02 moles = 0.001V moles
Solving for V, we find V = 20.
Hence, the volume of the HNO3 solution is 20 L.
Now that we know the volume of the HNO3 solution is 20 L, we can calculate the concentration of H+ ions:
Concentration of H+ ions = Moles of H+ ions / volume of solution
= 0.001V moles / V L
= 0.001 moles per liter
Converting this concentration to pH:
pH = -log[H+]
= -log(0.001)
= -(-3)
= 3
So, the pH of the final solution is 3.