Consider 0.40 M NaOH at 25{eq}^o {/eq}C. Determine {eq}~[OH^-] {/eq} and {eq}~[H_3O^+] {/eq} of this solution.
Question:
Consider 0.40 M NaOH at 25{eq}^o {/eq}C. Determine {eq}~[OH^-] {/eq} and {eq}~[H_3O^+] {/eq} of this solution.
Autoionization of Water:
Water is considered as a very weak electrolyte. Two water molecules one acting as an acid and other as a base react to produce hydronium ion in the solution.
{eq}H_2O (l) +H_2O(l) \leftrightharpoons H_3O^+ + OH^- {/eq}
The autoionization constant of water for every solution is constant at {eq}25^oC {/eq} and is given by {eq}K_w =\displaystyle {[H_3O^+ ][OH^-]} = 1 \times 10^{-14} {/eq}
Answer and Explanation: 1
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View this answerThe concentration of the NaOH solution is {eq}[NaOH] = 0.40 \ M {/eq}.
As NaOH is a strong base we can say that it is completely dissociated. Thus...
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Chapter 11 / Lesson 11Learn about acid-base indicators. Understand what acid-base indicators are, how they work to determine if a solution is acidic or basic, and see the examples.