# A solution with a pH of 7 has a [OH1-] of

## pH + pOH = pKw = 14. You know pKw and pH, solve for pOH.

Then pOH = -log(OH^-)

Solve for OH^-

## 7M. Is that correct?

## no. pOH is 7 (but not molar).

pH + pOH = 14

pH = 7; therefore, pOH = 7

Now do pOH = -log(OH^-) and solve for OH^-

## 1 x 10-7 M?? Is that correct?

## To determine the [OH-] concentration of a solution with a pH of 7, we can use the equation:

pOH = 14 - pH

Since the pH of the solution is 7, we can substitute this value into the equation:

pOH = 14 - 7

pOH = 7

Next, we can convert the pOH value to the [OH-] concentration:

[OH-] = 10^(-pOH)

Substituting the value of pOH into the equation:

[OH-] = 10^(-7)

Using a calculator, we can find that 10^(-7) is equal to 0.0000001.

Therefore, a solution with a pH of 7 has an [OH-] concentration of 0.0000001 (or 1 x 10^(-7)) moles per liter.