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If #pOH=10.75#, what is the concentration of #[HO^-]# in the solution?

Answer 1

Cameron Andrews

#[HO^-]=1.78xx10^-11*mol*L^-1.#

#pOH=-log_10[HO^-]#

If #pOH=10.75#, then #[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1.#

What are #pH#, and #[H_3O^+]# in this solution?

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Answer 2

Naomi Aronson

To find the concentration of ([HO^-]) in the solution, you can use the relationship:

[pOH = -\log[OH^-]]

Rearranging the equation gives:

[ [OH^-] = 10^{-pOH}]

Plugging in the given value of (pOH = 10.75):

[ [OH^-] = 10^{-10.75}]

[ [OH^-] = 1.778 \times 10^{-11}]

Therefore, the concentration of ([HO^-]) in the solution is (1.778 \times 10^{-11}) M.

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Answer 3

Xavier Avery

To find the concentration of ( \text{[HO}^-] ) in the solution, you can use the relation between ( \text{pOH} ) and ( \text{[OH}^-] ) concentration. The relation is given by:

[ \text{pOH} = -\log_{10}(\text{[OH}^-]) ]

Given ( \text{pOH} = 10.75 ), we can rearrange the equation to solve for ( \text{[OH}^-] ):

[ 10.75 = -\log_{10}(\text{[OH}^-]) ]

Now, we can solve for ( \text{[OH}^-] ):

[ \text{[OH}^-] = 10^{-10.75} ]

[ \text{[OH}^-] \approx 1.778 \times 10^{-11} ]

So, the concentration of ( \text{[HO}^-] ) in the solution is approximately ( 1.778 \times 10^{-11} ) moles per liter.

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Answer from HIX Tutor

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.