How many milliliters of 2.00 M HCl must be added to neutralize the following solutions? (a) A mixture of 0.160 M HNO3 (100.0 mL) and 0.100 M KOH (400.0 mL)

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Accepted Answer

Welcome back, everyone. We need to determine the volume of 3.25 molar nitric acid and milliliters required to neutralize a mixture of 350 mL of 2.25 molar hydrochloric acid and 300 mL of three molar sodium hydroxide. We're going to begin by writing out a balanced chemical equation. So we have hydrochloric acid reacting with sodium hydroxide in our mixture and recall that this type of reaction is known as a neutralization in which our products are going to be a salt and water are salt would be our sodium chloride, which is acquis and water is H Based on all of our coefficients. We can see that we actually can confirm that this reaction is balanced as written. Now, we need to determine which reaction or which reactant between hydrochloric acid and sodium hydroxide. We have in excess note that our ratio between hcl and sodium hydroxide is a 1-1 molar ratio based on our coefficients. And so recall that to get the moles of our reactant, we can find that by taking their polarity or molar concentration given in the prompt multiplied by the volume used where we would recall that molar concentration can be interpreted as equivalent to moles per leader. And so beginning with finding our moles of hydrochloric acid, we would take the polarity of hydrochloric acid multiplied by its volume In which were given in the prompt, the polarity as 2.25 Molar which we're interpreting as moles per liter. And we're going to multiply by the volume used as ml. But because we are given or we are interpreting our molar concentration as moles per liter, we want to convert this from male leaders to leaders. And so we would take 350 MLS and multiply by milliliters in denominator and leaders in the numerator where our prefix milli tells us that we have to the negative third of our base unit leader. And so canceling out milliliters, we get our volume of our acid being 0.35 liters, which we can multiply here below. And so this will allow us to cancel out leaders leaving us with moles of our hydrochloric acid in which we'll find we have a value of 0.7875 moles of HCL. And so now to find our moles of sodium hydroxide, our second reactant, we would again take its molar concentration multiplied by its volume. And so we're given a concentration from the prompt as three molar, which will interpret us three moles per liter. And then we want to multiply by its volume used being 300 mL, which we will convert to leaders in the same way by multiplying by 10 to the negative third power leaders. And we'll find that this results in 0.3 liters. So we would multiply by that factor. And so canceling out leaders were left with moles again, we'll find that we have In value of 0.9 moles of our sodium hydroxide. And because we see that 0.9 is much larger than .7, 8, 75, we would say therefore, sodium hydroxide is in excess and hydrochloric acid is our limiting reactant. Now, we need to determine our moles of excess sodium hydroxide. And so we can take the difference between .9 moles what we have in excess minus what we have limiting being our hydrochloric acid as .78-75 moles. And just to be clear, the .9 moles is moles of sodium hydroxide. So this difference results in our excess moles of sodium hydroxide being 0.1, 1, 25 moles. And so based on what our prompt tells us as far as determining the volume of nitric acid that is used to neutralize our mixture of HCL and sodium hydroxide. We can understand that. Therefore, based on this amount, our excess sodium hydroxide is neutralized By our nitric acid HN 03. So what we calculated 0.1125 moles of sodium hydroxide is what is the amount neutralized by sodium hydroxide? And so now we want to write out us an equation to represent this where we have sodium hydroxide reacting to neutralize nitric acid or sorry, nitric acid neutralizing our sodium hydroxide. And in this reaction, we form yet again, another neutralization reaction in which we have assault as our first product being sodium nitrate and water. Of course, as our second product because we can see based on our coefficients and number of atoms, we have definitely another bounced reaction. As it's written, we can say that our moles of excess based sodium hydroxide are equal to our moles of nitric acid which we are neutralizing, which we are using to neutralize. And so we can understand that based on our calculated excess moles of sodium hydroxide, we have for moles of nitric acid, An amount equal to 0.1, 1, 25 moles. But we need our volume of nitric acid to neutralize our base. And so we're going to recall that we can take the volume of nitric acid or we can find the volume of nitric acid by taking our moles of nitric acid divided by the molar itty of nitric acid which we are given in the prompt. And so are moles of nitric acid. We just determined equivalent to our excess space as 0.1125 moles is our numerator And our molar concentration going back to our prompt for nitric acid is given as 3.25 molar. So we would plug that in our denominator which we will interpret as 3.25 moles per liter. And so you can see we can cancel our units of moles in the numerator with moles in the denominator, leaving us with leaders in the denominator. However, we want our volume to be in mill leaders since the volume is given in the prompt for the rest of our re agents are also in middle leaders. And so we're going to multiply by the conversion factor where we understand that for one leader and this will actually be in the denominator here. We have an equivalent of 10 to the third-power mill leaders. Now because we canceled out moles were technically left with leaders in the numerator or rather not even in a fraction anymore, which is why we can actually cancel it out with the denominator of leaders over here. So we can cancel out leaders and we're left with male leaders as our final unit of volume. And so in our calculators, the product here will result in a value of 34. ml. And this would be our final answer as our volume of nitric acid that is required to neutralize our mixture of hydrochloric acid and sodium hydroxide. This corresponds to choice d in the multiple choice as our final answer, I hope everything I went through is clear if you have any questions, please leave them down below and I'll see everyone in the next practice video.

How many milliliters of 2.00 M HCl must be added to neutralize the following solutions? (a) A mixture of 0.160 M HNO3 (100.0 mL) and 0.100 M KOH (400.0 mL)

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Key Concepts

Neutralization Reaction

Molarity (M)

Stoichiometry