What will be the pH of an aqueous solution containing 0.040 mol/L sodium hydroxide??
I got 1.40 or 1.398 I rounded the numbers is
my answer right??? If not what's the right answer??
I know I wrote the question again, because on the previous answer I did not understand what u mean and from where u got 14.00 ... Plz can u explain it again so that I get it . It's very important to me thank u :)
NaOH ==> Na^+ + OH^-
NaOH is 0.04M from the problem.
Na^+ is 0.04M and OH^- is 0.04M
You took the -log(OH^-), which is 1.40 and you called that the H^+ but it isn't. It is the (OH^-) so you calculated the pOH and not the pH. But you can convert your pOH to pH from the equation
pH + pOH = pKw = 14.00
pH + 1.40 = 14.00
pH = 14.00 - 1.40 = 12.60
To calculate the pH of a solution, you need to use the formula:
pH = -log[H+]
In this case, we will determine the concentration of hydroxide ions (OH-) first, as sodium hydroxide (NaOH) is a strong base that dissociates completely in water.
The concentration of hydroxide ions in a 0.040 mol/L sodium hydroxide solution is also 0.040 mol/L because sodium hydroxide dissociates to produce one hydroxide ion for every one mole of sodium hydroxide.
Now, to find the concentration of hydrogen ions (H+), we use the concept of Kw, which is the ion product of water at a given temperature. At 25°C, Kw is equal to 1.0 x 10^-14.
In a neutral solution, the concentration of hydrogen ions (H+) is equal to the concentration of hydroxide ions (OH-), and thus both concentrations are 1.0 x 10^-7.
Since this is a basic solution, we use the equation:
Kw = [H+][OH-]
Therefore, [H+] = Kw / [OH-]
= (1.0 x 10^-14) / (0.040)
= 2.50 x 10^-13 mol/L
Finally, we can calculate the pH using the formula:
pH = -log[H+] = -log(2.50 x 10^-13) = 12.60
So, the correct answer for the pH of the aqueous solution containing 0.040 mol/L sodium hydroxide is pH = 12.60, which is different from your previous answers of 1.40 and 1.398.
To determine the pH of a solution, you need to use the formula: pH = -log[H+], where [H+] represents the concentration of hydrogen ions in the solution.
In the case of sodium hydroxide (NaOH), it is a strong base that dissociates completely in water to release hydroxide ions (OH-). Since NaOH is completely ionized, the concentration of hydroxide ions is the same as the concentration of sodium hydroxide.
So, in this case, the concentration of hydroxide ions is 0.040 mol/L. Now, to find the concentration of hydrogen ions, you can use the fact that in water, the product of hydrogen ion concentration and hydroxide ion concentration is constant and equal to 1 x 10^-14.
[H+][OH-] = 1 x 10^-14
Since the concentration of hydroxide ions is known to be 0.040 mol/L, you can rearrange the equation and calculate the concentration of hydrogen ions:
[H+] = (1 x 10^-14) / 0.040
[H+] = 2.5 x 10^-13 mol/L
Now, to find the pH, take the negative log of the hydrogen ion concentration:
pH = -log(2.5 x 10^-13)
Calculating this, you will get a pH of approximately 12.60.
So, the correct answer for the pH of the solution containing 0.040 mol/L sodium hydroxide is 12.60.