(c) If 18.65 mL
of the caesium hydroxide solution was needed to neutralize a
42.3 mL aliquot of the hydroiodic acid solution, what is the
concentration (molarity) of the acid?

Question

Accepted Answer

Hey everyone, we're asked to calculate the molar concentration of Hydro chaotic solution. If 22.4 ml of 0.915 rubidium hydroxide was required to completely neutralize our acid. So first let's go ahead and write our reaction. We have our hydro biotic solution and we react this with our rubidium hydroxide. Since we have an acid based reaction, we're going to end up with rubidium, I died plus water and checking our reaction out, we can see that everything is balanced so we can go ahead and move on with our question. Starting off with our 22.4 ml of rubidium hydroxide. Since this is what was required to neutralize our asset, We're going to convert this into leaders and we know that 10 to the 3rd mill leaders is equivalent to one leader And we were told that our polarity of rubidium hydroxide was 0. moles of rubidium hydroxide per one leader. Next looking at our multiple ratios between rubidium hydroxide and our hydra idiotic acid, We see that we have a 1-1 ratio And this will get us to a value of 0. 96 mol of hydro ionic acid. Now to calculate our polarity we can go ahead and take our 0.020496 mol of hydrochloric acid. And we can divide this by our 51.4 milliliters. And since we want this in leaders for polarity, we're going to use our dimensional analysis and we know that we have 10 to the third middle leaders per one leader. And this will get us to a value of 0. Mueller. And this is our final answer. So I hope this made sense and let us know if you have any questions.

(c) If 18.65 mL of the caesium hydroxide solution was needed to neutralize a 42.3 mL aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

Verified Solution

Key Concepts

Neutralization Reaction

Molarity

Stoichiometry