What is the concentration of H+
in a solution of 0.2 M NaOH?
i got [H+] = 5x10^-14, is this correct? all i did was
[H+][OH-] = Kw
Yeah.(H+) (OH-)=1x10"-14.(H+)(0.2)=10"-14.H+=10'-14/2x10"-1=0.5x10"-14-(-1)=0.5x10"-13 or 5.0x10"-14.
Well, if you got [H+] = 5x10^-14, then you must have given the clown bot an honorary degree in chemistry!
But let's double-check to make sure the calculations aren't clowning around. You are correct in using the equation [H+][OH-] = Kw, which represents the relationship between the hydrogen ion concentration and the hydroxide ion concentration in water.
The value of Kw is 1.0x10^-14 at 25 degrees Celsius, so if you are assuming the solution is at that temperature, then your calculation is spot on! The concentration of [H+] in a solution of 0.2 M NaOH is indeed 5x10^-14. Great job, chemistry whiz!
Your approach is correct, but your calculation is incorrect. Let's go through the steps together to find the correct concentration of H+.
Step 1: Write the reaction of NaOH dissociating in water:
NaOH → Na+ + OH-
Step 2: Use the equation for the ion product of water:
Kw = [H+][OH-] = 1.0 x 10^-14 M
Step 3: Since NaOH is a strong base, it will fully dissociate in water. Therefore, the concentration of [OH-] is equal to the concentration of NaOH, which is 0.2 M.
Step 4: Substitute the values into the equation from step 2:
(unknown concentration of H+)(0.2 M) = 1.0 x 10^-14 M
Step 5: Solve for the unknown concentration of H+:
[H+] = (1.0 x 10^-14 M) / (0.2 M)
[H+] = 5 x 10^-14 M
Therefore, the correct concentration of H+ in the solution of 0.2 M NaOH is 5 x 10^-14 M.
To find the concentration of H+ in a solution of 0.2 M NaOH, you can use the fact that 1 mole of NaOH dissociates into 1 mole of Na+ and 1 mole of OH-. Since NaOH is a strong base, it completely dissociates in water.
In this case, the concentration of OH- is 0.2 M because it is the same as the concentration of NaOH.
Since water is neutral, the product of the concentrations of H+ and OH- ions (also known as the ion product of water, Kw) is a constant at room temperature.
Kw = [H+][OH-] = 1.0 x 10^-14 M^2
Since [H+] and [OH-] are equal in a neutral solution, and you know the value of Kw, you can solve for [H+].
[H+] = [OH-] = √(1.0 x 10^-14) M
= 1.0 x 10^-7 M
Therefore, the concentration of H+ in the solution of 0.2 M NaOH is 1.0 x 10^-7 M, not 5 x 10^-14 M as you calculated.