Answer :

Final answer:

The molarity of Sr(OH), is 2 * 10^(-1.056).

Explanation:

Since the pH is a measure of the concentration of H+ ions, we can use the formula:

pOH = 14 - pH

where pOH is the negative logarithm of the hydroxide ion concentration. In this case, the pOH is:

pOH = 14 - 12.944

= 1.056

Now, we can use the relationship between pOH and the concentration of hydroxide ions to find the concentration of OH- ions:

pOH = -log[OH-]

where [OH-] is the concentration of hydroxide ions. Rearranging the equation, we get:

[OH-] = 10^(-pOH)

Substituting the value of pOH, we find:

[OH-] = 10^(-1.056)

Finally, to find the molarity of Sr(OH),, we need to use stoichiometry. The balanced chemical equation for Sr(OH), is:

Sr(OH), -> Sr^(2+) + 2OH-

From the equation, we can see that for every Sr(OH), molecule, we get 2 OH- ions. Therefore, the molarity of Sr(OH), is twice the concentration of OH- ions:

Molarity of Sr(OH), = 2 * [OH-]

Learn more about calculating molarity of a solution here:

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To calculate the molarity of Sr(OH),, we first need to convert the pH to the concentration of hydroxide ions (OH-). The pH scale is a logarithmic scale that measures the concentration of hydrogen ions (H+) in a solution. A pH of 12.944 indicates a high concentration of hydroxide ions (OH-) in the solution.

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