Question

What is the pH of a 6.50 x 10-3 M KOH solution?

Answer 1

Answer:

11.81

Explanation:

We'll begin by writing the balanced dissociation equation for KOH. This is given below:

KOH —> K+ + OH-

Next, we shall determine the concentration of Hydroxide ion, [OH-]. This is illustrated below:

From the balanced equation above,

We can see that 1 mole of KOH produced 1 mole of OH-.

Therefore, 6.50x10^-3 M KOH will also produce 6.50x10^-3 M OH-.

Next, we shall determine the pOH of the solution. This is illustrated below:

pOH = –Log [OH-]

[OH-] = 6.50x10^-3 M

pOH = –Log 6.50x10^-3

pOH = 2.19

Finally, we can determine the pH of the solution as follow:

pH + pOH = 14

pOH = 2.19

pH + 2.19 = 14

Collect like terms

pH = 14 – 2.19

pH = 11.81

Therefore, the pH of the solution is 11.81

Answer 2

The pH of a 6.50 x 10-3 M KOH solution is 11.81 which is alkaline.

pH is a measure of how acidic or basic a solution is. Ph has a  range from 0 to 14. pHs with  less than 7 show  acidity, while a pH of greater than 7 indicates a base/ alkalinity. , pH of  7 shows the solution is neutral.

KOH is a strong base, Upon dissociation,it gives

KOH -> K⁺ + OH⁻

Here, 1 mole of KOH produced 1 mole of OH-.

Therefore, 6.50 x 10-3 M KOH will prodce 6.50 x 10-3 M H

Now, we first determine the pOH of the solution with the equation:

pOH = –Log [OH-]

[OH-] = 6.50x10^-3 M

pOH = –Log 6.50x10^-3

pOH = 2.19

Now, we know that  

pH + pOH = 14

Having gotten that  

pOH = 2.19,

pH + 2.19 = 14

pH = 14 – 2.19

pH = 11.81

Therefore, the pH of the KOH solution is 11.81 which is alkaline.

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