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# Strong base solutions

Strong bases (such as Group 1 and 2 metal hydroxides) dissociate completely in water to produce hydroxide ions. The concentration of OH⁻ in a strong base solution can therefore be determined from the initial concentration of the base and the stoichiometry of the dissolution. For example, the strong base Ba(OH)₂ dissociates to give two OH⁻ ions per formula unit, so a 0.1 M Ba(OH)₂ solution has an OH⁻ concentration of 0.2 M. Created by Jay.

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• At , how come the final concentration for OH- is to two significant figures, when all inputs were to three significant figures? • Different mathematical operations have their own rules for sig figs. Here Jay is using logs and antilogs which have different rules compared to multiplication and division. Antilogs, which the last operation Jay uses before he gets the hydroxide concentration, states that an answer should have as many sig figs as the mantissa (the decimal portion) of the number being antiloged. So if we're taking 10^(-1.00) and the mantissa are the two 0s then the answer should only have two sig figs then.

Hope that helps.
• I don't quite understand at why all of the hydroxide ions come from the strong base CaOH2. I mean, doesn't the autoionization of water also play a role in creating OH-? It seems to me that we are underestimating the concentration of OH-
(1 vote) • Technically the hydroxide concentration is due to both the calcium hydroxide and the autoionization of water. However, they have vastly different contributions. The calcium hydroxide delivers a hydroxide concentration of 0.0020 M, while water’s autoionization provides only 1.0 x 10^(-7) or 0.00000010 M. So 0.0020 M + 1.0x10^(-7) M = 0.0020007 M. And when we take the pOH it becomes 2.698818027861, which rounded for sig figs still becomes 2.70 like the pOH in the video. So since water’s autoionization contribution is so small that it’s not even noticeable in the final answer, it’s simply omitted since its inclusion is insignificant compared to calcium hydroxide’s contribution.

Hope that helps.
• Why do they find the pOH using the concentration of the OH and then subtracting it from 14 instead of just using the Ca to find pH?

thank you to anyone who answers this
(1 vote) • Calcium doesn’t affect the pH of the solution, only the hydroxide does. This is because hydroxide reacts with hydrogen ions, or more accurately hydronium, and neutralizes into water. As a result, hydroxide decreases the concentration of the hydronium and raises the pH. We can relate the hydronium concentration and pH through the autoionization of water.

The autoionization of water is: 2H2O(l) ⇌ H3O^(+)(aq) + OH^(-)(aq)
Which means we can write the equilibrium expression as: Kw = [H3O^(+)][OH^(-)]
And if we apply –log to both sides the equilibrium expression becomes: pKw = pH + pOH. Where pKw is 14 at 25°C.

So if we don’t know the hydronium concentration directly, we can still calculate it and the pH through the hydroxide concentration and pOH. Knowing the concentration of calcium doesn’t help us find the hydronium concentration or pOH, at best it allows us to indirectly know the hydroxide concentration.

Hope that helps.  