How many milliliters of 1.00 M KOH must be added to neutralize th...

Question

How many milliliters of 1.00 M KOH must be added to neutralize the following solutions?
(a) A mixture of 0.240 M LiOH (25.0 mL) and 0.200 M HBr (75.0 mL)

Accepted Answer

Hey everyone. Our question here wants us to determine the volume of chronic acid used in this solution 1st. Let's go ahead and write out our reaction. So we have rubidium hydroxide and it reacts with chronic acid. Since we have an acid-based reaction, we end up with water plus rubidium chrome eight. Now, before we move on with our question, let's first balance out our reaction. As we can see in our product side, we have two of rubidium so we'll need to add a coefficient of two prior to rubidium hydroxide And since we adjusted our oxygen and hydrogen, we need to add a coefficient of two as well prior to our water. And now our reaction is completely balanced. Now starting off with our 0.424g of rubidium hydroxide, We're going to take the molar mass of rubidium hydroxide which is 102.475 g of rubidium hydroxide per one mole. And we'll go ahead and take our multiple ratios. And as we can see we have two moles of rubidium hydroxide per one mole of chronic acid. And since we want to end up in mill leaders of our chronic acid Will take the polarity of chronic acid which is 0.310 moles of chronic acid per one leader. And to convert this into middle leaders we know that one leader contains 10 to the 3rd mill leaders. And when we calculate this out and cancel out all our units, we end up with a value of 6.67 mill leaders of chronic acid and this is going to be our final answer. So I hope this made sense and let us know if you have any questions.

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