Consider a 500mL solution containing 2.0×10^-5 mol HCl
·what is the pH of the above solution?
·If the solution is diluted to 1.5 L what will be the new pH?
conc HCL=2.0E-5/2=1E-5
pH=-log concH+=5
now, diltued 3x as above
pH=-log1/3 E-5=- log .333E-5=5.48
2 x 10⁻⁵mole HCl in 500-ml Solution:
=>[HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liters)
= (2 x 10⁻⁵ mole H⁺/ 0.500 Liters) = 4.0 x 10⁻⁵ Molar in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(4 x 10⁻⁵) = 4.4
Diluting to 1.5 Liters:
=> [HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liter) = (2 x 10⁻⁵ mole H⁺/ 1.500 Liters) = 1.33 x 10⁻⁵M in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(1.33 x 10⁻⁵) = 4.9
Dr Rebel is correct, I divided 2 by .5 and got 1. UGH
To find the pH of a solution, we need to know the concentration of H+ ions in the solution. In the case of a strong acid like HCl, the concentration of H+ ions is equal to the concentration of HCl. Therefore, we can use the given concentration of HCl to determine the pH.
1. Calculate the concentration of HCl:
The molarity of a solution (M) is defined as the number of moles of solute per liter of solution. Since we have a 500 mL (0.5 L) solution containing 2.0×10^-5 mol of HCl, we can calculate the concentration (C) as follows:
C = (number of moles of HCl) / (volume of solution in liters)
C = (2.0×10^-5 mol) / (0.5 L)
C = 4.0×10^-5 M
2. Convert the concentration to pH:
pH is a measure of the acidity or alkalinity of a solution and is defined as the negative logarithm (base 10) of the H+ concentration. The formula for finding pH is:
pH = -log[H+]
Since the concentration of H+ ions is equal to the concentration of HCl, we can write:
[H+] = 4.0×10^-5 M
Substitute this value into the pH formula:
pH = -log(4.0×10^-5)
pH ≈ 4.40
Therefore, the pH of the 500 mL solution containing 2.0×10^-5 mol of HCl is approximately 4.40.
Now, let's move on to the second part of the question:
To find the new pH after dilution, we need to calculate the new concentration of the HCl solution after it is diluted to 1.5 L. Dilution does not affect the pH of the solution since the concentration of H+ ions remains the same.
3. Calculate the new concentration after dilution:
The general equation for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, we have the initial concentration (C1 = 4.0×10^-5 M), initial volume (V1 = 0.5 L), and final volume (V2 = 1.5 L). Let's solve for C2:
C1V1 = C2V2
(4.0×10^-5 M)(0.5 L) = C2(1.5 L)
2.0×10^-5 mol = C2(1.5 L)
Divide both sides by 1.5 L to solve for C2:
C2 = (2.0×10^-5 mol) / (1.5 L)
C2 = 1.33×10^-5 M
Therefore, after diluting the solution to 1.5 L, the new concentration of HCl is approximately 1.33×10^-5 M. Since the concentration of H+ ions remains the same, the new pH of the diluted solution is still approximately 4.40.