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What are the hydronium and hydroxide ion concentrations in a 3 x 10-2 M NaOH solution? [ H 3 O +...

Question:

What are the hydronium and hydroxide ion concentrations in a 3 x 10-2 M NaOH solution?

{eq}[H3O+] = 3.3 \times 10^{-13} M, [OH-]= 1 \times10^{-2} M\\ [H3O+] = 3.3 \times 10^{-13} M, [OH-]= 3.3 \times10^{-13} M\\ [H3O+] = 3 \times 10^{-2} M, [OH-]= 3.3 \times10^{-13} M\\ [H3O+] = 3.3 \times 10^{-13} M, [OH-]= 3 \times10^{-2} M {/eq}

When I did {eq}1\times10^{-14}/3\times10^2 {/eq}, I got {eq}3.33\times10^{-17} {/eq} on my calculator as well as online, which is not an option.

Answer and Explanation:

The problem at our hand is very trivial provided we have our basics about ionic concentrations in place, not to forget the basics of mathematics too. To be able to crack this up all we need to know is the given equation:
{eq}[H^+][OH^-] = 10^{-14} {/eq}
What above equation describes is the concentrations relationship between the hydronium and the hydroxide ions at any point in time. With given information that the concentration of the NaOH solution is {eq}3 \times 10^{-2} {/eq} M, and the knowledge that NaOH is a strong base meaning it dissociates completely, it is safe to conclude that the concentration of the hydroxide in the solution would be:
{eq}[OH^-] = 3 \times 10^{-2} {/eq} M. With this much sorted, we now use the equation above to determine the hydronium ion concentration:
{eq}[H^+][OH^-] = 10^{-14} \\ [H^+]\times(3 \times 10^{-2}) = 10^{-14} \\ [H^+] = 10^{-14} / (3 \times 10^{-2}) = 3.3 \times 10^{-13} M {/eq}
So the answer is option (d).
Taking a closer look at your calculation, you would see that you plugged in {eq}3 \times 10^{2} {/eq} rather than {eq}3 \times 10^{-2} {/eq}, omitting the negative sign in the power, and thus the incorrect answer.


Learn more about this topic:

Ionic Equilibrium: Definition & Calculations

from UPSEE Paper 1: Study Guide & Test Prep

Chapter 24 / Lesson 3
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