A solution is prepared by pipetting 5.00 mL of 0.0983 molar HCl into a 100.0 mL graduated cylinder, and adding enough water to prepare 50.0 mL of solution.
1.The concentration of this solution is ____________M
2.The pH of this solution is ____________
Because HCl is a strong acid therefore it will dissolve 100% therefore H=0.0983M and Cl =0.0983
pH=-log(0.0983)
I have this but it not correct.
To find the concentration of the solution, you need to use the dilution formula, which states that the initial concentration multiplied by the initial volume equals the final concentration multiplied by the final volume.
In this case, the initial concentration (HCl) is 0.0983 M, and the initial volume is 5.00 mL. The final volume of the solution is 50.0 mL, as mentioned. We can set up the equation as:
0.0983 M x 5.00 mL = final concentration x 50.0 mL
Simplifying the equation, we can solve for the final concentration:
final concentration = (0.0983 M x 5.00 mL) / 50.0 mL
Calculating that, we get:
final concentration = 0.00983 M
So the concentration of the solution is 0.00983 M.
Now let's move on to calculating the pH. Since the solution is made from HCl, a strong acid, it will dissociate completely in water. The concentration of H+ ions will be equal to the concentration of HCl, which is 0.00983 M.
To calculate the pH, you correctly mentioned that pH = -log([H+]). So we can calculate it as follows:
pH = -log(0.00983)
Now let's calculate the pH value using a calculator:
pH ≈ 2.01
Therefore, the pH of this solution is approximately 2.01.
To find the concentration of the solution, you need to consider the final volume of the solution. The final volume of the solution is 50.0 mL.
Since you started with 5.00 mL of 0.0983 M HCl and then diluted it to a final volume of 50.0 mL, you can use the equation for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (0.0983 M)
V1 = initial volume of the solution (5.00 mL)
C2 = final concentration of the solution (unknown)
V2 = final volume of the solution (50.0 mL)
Rearranging the equation to solve for C2, we have:
C2 = (C1 * V1) / V2
Plugging in the values:
C2 = (0.0983 M * 5.00 mL) / 50.0 mL
C2 = 0.00983 M
Therefore, the concentration of this solution is 0.00983 M.
To find the pH of the solution, you can use the equation:
pH = -log[H+]
where [H+] is the hydrogen ion concentration, which is equal to the concentration of the HCl solution.
Using the given concentration of 0.0983 M:
pH = -log(0.0983)
Calculating this in a calculator, we get:
pH ≈ 1.006
Therefore, the pH of this solution is approximately 1.006.
(1)Ca. Moles H = Molarity x Volume(L) = 0.005L(0.0983M)HCl = 0.00049 mole HCl in 50ml of solution. Now, covert to [HCl] = (moles HCl)/(Vol. Soln in Liters) = 0.00049mole/0.050L =0.0098M in HCl => [H]=0.0098M.
(2) pH = -log[H] = -log(0.0098) = 2.01