# What is the H+ concentration (in molarity) of a solution with a pOH of 6.7?

The relationship that you wish to use is this one:

Given the pOH, we can calculate the pH by deducting it from 14 in the following manner:

14 - pOH is equal to 7.3.

I hope you can understand this explanation!

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To find the (H^+) concentration (([H^+])), you can use the relationship between (pOH) and ([OH^−]), and then use the fact that in a neutral solution, ([H^+] \times [OH^−] = 1.0 \times 10^{−14}) at 25°C. First, calculate ([OH^−]) using the formula (pOH = -\log[OH^−]), then find ([H^+]) by taking the reciprocal of ([OH^−]).

Given: (pOH = 6.7)

Calculate ([OH^−]): [ [OH^−] = 10^{-pOH} = 10^{-6.7} ]

Calculate ([H^+]): [ [H^+] = \frac{1.0 \times 10^{-14}}{[OH^−]} ]

[ [H^+] = \frac{1.0 \times 10^{-14}}{10^{-6.7}} ]

[ [H^+] = 10^{(-14 + 6.7)} ]

[ [H^+] = 10^{-7.3} ]

So, the (H^+) concentration of the solution is (10^{-7.3}) M.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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