A 200 mL sample of 1.0 M nitric acid is added to 1.0 M water. What is the resulting pH?
I think I'm doing the same lab as you. So what I did was find the new concentration of nitric acid in the solution (because by adding it to water, you've just diluted ur concentration). I got .167. Since nitric acid is a strong acid, it dissociates completely, right? So the H+ concentration is .167M and so you take the -log of that and get your pH. Does that make sense?
What is the volume of the solution?
1.2L is the volume
Well, when in doubt, always remember that water is a great liquid for a fish, but not always great for measuring pH! Now, let me try to clown-around with some numbers here. Since you're mixing equal volumes of two solutions with a concentration of 1.0 M, it's like throwing one clown and another clown into a big clown car. The total volume becomes 400 mL, but the concentration of nitric acid doesn't change. So, if we assume the pKa of nitric acid as approximately -1.4, we can estimate the pH to be somewhere around -0.7. But hey, take my answer with a pinch of salt and a dash of clown makeup, because it's just a wild guess!
To determine the resulting pH of the mixture, we need to consider the properties of the nitric acid (HNO3) and water (H2O) when they react together. Nitric acid is a strong acid, meaning it dissociates completely in water, while water is a neutral substance.
When nitric acid reacts with water, it donates a proton (H+) to the water molecules, resulting in the formation of hydronium ions (H3O+). This increases the concentration of hydronium ions, leading to a decrease in pH value.
Since the initial solutions of nitric acid and water are both 1.0 M, their concentrations will remain the same after mixing. The volume of the resulting solution is the sum of the individual volumes: 200 mL (nitric acid) + 200 mL (water) = 400 mL.
To calculate the new concentration of hydronium ions, we can use the equation:
[H3O+] = [acid] + [water]
[H3O+] = (volume of acid)(concentration of acid) + (volume of water)(concentration of water)
= (200 mL)(1.0 M) + (200 mL)(1.0 M)
= 200 mmol + 200 mmol
= 400 mmol
To find the resulting pH, we can use the formula:
pH = -log[H3O+]
pH = -log(400 x 10^-3)
= -log(4 x 10^-1)
= -log(4) + log(10^-1)
= -0.602 + 1
= 0.398
Therefore, the resulting pH of the mixture is approximately 0.398.