0.70 M KOH
find the pH
To find the pH of 0.70 M KOH, we need to calculate the hydroxide ion concentration (OH-) and then use it to find the pOH. Finally, we will use the pOH to calculate the pH.
1. Calculate the hydroxide ion concentration (OH-):
KOH is a strong base that dissociates completely in water, so 0.70 M KOH will produce 0.70 M OH- ions.
2. Calculate the pOH:
The pOH is the negative logarithm (base 10) of the hydroxide ion concentration (OH-).
pOH = -log[OH-]
pOH = -log(0.70) ≈ 0.155
3. Calculate the pH:
The pH is the negative logarithm (base 10) of the hydrogen ion concentration (H+).
pH + pOH = 14 (at 25°C)
pH = 14 - pOH
pH = 14 - 0.155 ≈ 13.845
Therefore, the pH of 0.70 M KOH is approximately 13.845.
To find the pH of a solution of potassium hydroxide (KOH) with a concentration of 0.70 M, you need to calculate the concentration of hydroxide ions (OH-) in the solution. Then, you can use the formula for pH to determine the acidity or basicity of the solution.
First, let's determine the concentration of hydroxide ions (OH-) in the solution. Since KOH is a strong base, it fully dissociates in water to produce one hydroxide ion for every potassium ion. Therefore, the concentration of hydroxide ions is the same as the concentration of KOH.
In this case, the concentration of KOH is given as 0.70 M. Therefore, the concentration of hydroxide ions (OH-) is also 0.70 M.
Now, to calculate the pH, we need to use the equation:
pOH = -log[OH-]
To convert pOH to pH, we can use the equation:
pH = 14 - pOH
First, let's calculate pOH:
pOH = -log[OH-]
pOH = -log[0.70]
Using a calculator, we find:
pOH ≈ 0.155
Now, let's calculate the pH:
pH = 14 - pOH
pH = 14 - 0.155
Using a calculator, we find:
pH ≈ 13.845
Therefore, the pH of a 0.70 M KOH solution is approximately 13.845.
Since KOH is a strong base, it totally dissociates, and thus the concentration of hydroxide ions in solution equals to the concentration of the KOH.
0.70M KOH
note that:
[H+][OH-]= 1.0 x 10^-14
so,
[H+]= 1.0x10^-14 / [0.70M] = ______[H+]
since pH = -log[H+]
plug in the [H+] you found above into the equation for pH.