Calculate the pH of a solution prepared by diluting 0.0015L of a 0.100 mol/L nitric acid solution with 0.5L of water.
*C1×V1=C2×V2
C2=0.3MOL/L
Do we use this concentration to find the pH
Your use of c1v1 = c2v2 is ok but I don't think you punched in the right numbers. Or set it up wrong.
0.1 x 0.0015 = C2 x 0.5
C2 = ?
And yes, you use C2 to solve for pH. I obtained about 3.5(estimated)
No, in this case, we need to use the concentration before the dilution (C1) to find the pH. The formula you mentioned, C1×V1=C2×V2, is the dilution formula, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume after dilution.
To find the pH of the solution, we need to know the pKa of nitric acid, which represents the dissociation constant for the acid. However, since nitric acid (HNO3) is a strong acid, it dissociates completely in water, meaning it has a pKa value close to -1. So, we can use the concentration of the original nitric acid solution, C1, to calculate the pH.
To find the pH, we use the formula:
pH = -log[H+]
Since nitric acid is a strong acid, it fully dissociates into H+ ions in water. Therefore, the concentration of H+ in the solution will be the same as the initial concentration of nitric acid (C1).
In this case, the initial volume of the nitric acid solution is 0.0015 L, and the initial concentration is 0.100 mol/L. Multiplying these values gives us:
[H+] = C1 = 0.100 mol/L
Taking the negative logarithm of 0.100 gives us:
pH = -log(0.100)
Using a calculator, we find:
pH ≈ 1
Therefore, the pH of the diluted solution prepared by diluting 0.0015 L of a 0.100 mol/L nitric acid solution with 0.5 L of water is approximately 1.