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Question 414 pts20.0 ml of HPzO6 solution is neutralized by 40.0 ml of 0.800 M KOH_ Calculate the molarity "M" of the HaPzOs solution:*Submit scratch pape...

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Question 414 pts20.0 ml of HPzO6 solution is neutralized by 40.0 ml of 0.800 M KOH_ Calculate the molarity "M" of the HaPzOs solution:*Submit scratch paper showing set-Up, numerical calculations and answer:

Question 4 14 pts 20.0 ml of HPzO6 solution is neutralized by 40.0 ml of 0.800 M KOH_ Calculate the molarity "M" of the HaPzOs solution: *Submit scratch paper showing set-Up, numerical calculations and answer:



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If $38.2 \mathrm{~mL}$ of a $0.163 \mathrm{M}$ KOH solution is required to titrate $25.0 \mathrm{~mL}$ of a solution of $\mathrm{H}_{2} \mathrm{SO}_{4},$ what is the molarity of the $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution? $$\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)$$

More clarity is defined as moles per liter. So to calculate the polarity of the sulfuric acid solution, we need to know the moles of sulfuric acid, which we can calculate knowing the stock geometry of the reaction and the amount of potassium hydroxide or sodium hydroxide Looking like it was switched in this problem required to react with the sulfuric acid and then we divided by the volume of sulfuric acid in leaders, which was provided 25 ml. Here's the balanced chemical reaction. There's a 2-1 relationship between sulfuric acid and the base with which is just reacting. We'll start with the 32.8 million lit of sodium hydroxide or potassium hydroxide converted into Leaders by dividing by 1000. Then when we have leaders of the based solution will multiply it by the concentration of the based solution in order to get moles of the base. Then we recognize that two moles of the base are required for every one mole of the acid. This will give us moles of the h two s 04 We then divide that by the volume of the H two s 04 solution in leaders. It was provided at 25 mil leaders, who will divide that by 1000 to get units of leaders. This then gives us 10000.1 oh six moller H two s 04

Dr 15 Problem 25 asks us to determine the polarity of a k o H solution if we're given the amount of H two s 04 required to neutralize it. So first, let's start by writing out are balanced chemical equation here. So we have hte to s 04 and we're reacting it with ko each. Now remember that every acid base neutralization reaction will give us water as one of the products and assault as the other product. So here that salt is K s 04 Now, to balance this out will need to put a to in front of K O. H and in front of water. So notice here that h two s 04 gave us two hydrogen ins per molecule and K o. H gave us 10 h per molecule. So we needed twice a much k o. H or base as our asset. Now we can use the volume and the polarity that was given to us for our acid and use that to calculate the molds and eventual polarity of our base. So we're told that we had 2.5 times 10 to the negative two moles or moles per leader. So if we multiply that by the number of leaders or 0.15 that will give us the moles of our acid, which here is your a 0.375 Mol of H two s 04 Now we can use the ratios from our balanced chemical equation to convert this two moles of K o h. So for every Mol of h two s 04 it's two moles of K O h, which gives us 0.75 mol of K a witch. Finally, we can convert moles to polarity by dividing by leaders so 00075 Mol divided by 0.1 leaders gives us 0.75 Moeller que O h

To calculate the pH after particular volumes of the strong base potassium hydroxide has been added to 40 mil. Leaders of a point to Moller solution of strong acid hcl 04 We need to first figure out what the equivalence point volume is so we can make sense of the five volumes that are given to us in this particular problem. To calculate the equivalence point volume, we will first start with the volume of hi Paul uh, per caloric acid HCL 04 which is 40 mL converted into leaders, then used the molar ity of per clark acid at 400.2 Moeller to convert the leaders into moles of per clark acid, then recognize that the reaction is one toe one. So one mole of per clark acid reacts with one mole of potassium hydroxide. Then when we have moles, potassium hydroxide, we can use the mill arat e of the potassium hydroxide solution to calculate the leaders of potassium hydroxide that will neutralize the 40 million leaders of per caloric acid and then convert the leaders into milliliters and the equivalents point volume, then, is 80 mL of potassium hydroxide. So the first volume at zero mill leaders. The pH of the solution is just going to be a function of the concentration of the strong acid. The strong acid concentration will be the hydro knee. Um, concentration. We take the negative log of that in order to get the pH at 00.699 after 10 mil. Leaders of potassium hydroxide has been added. A corresponding amount of perk Lorik acid in moles will have reacted. So to solve for the hydro knee, um, concentration, we need to take the moles of per caloric acid that we start with, which will be the 40 mil leaders converted to leaders at 400.4 multiplied by the concentration. This is the moles per caloric acid that we start with within. Subtract off the moles of folkloric acid that reacted which will be equal to the moles of potassium hydroxide added which is 10 mL at 100.1 Mueller And then we divide by the new volume as we have increased the volume from 40 mL 2 50 mL with the addition of 10 mL. Then when we know the hydro knee, um, concentration, we can take the negative log to get a pH of 0.854 after 40 mL have been added, we have reached the half equivalents point hydro knee. Um, concentration will be calculated the same way as we did appear because we still have excess hydro knee. Um, we have not yet reached the equivalence point. So we have the moles of perk Lorik acid that we started with which will be equal to the moles of hydro knee. Um, that we start with subtracting off the moles of potassium hydroxide that has been added 40 mL, converted to leaders multiplied by its concentration of 400.1. And then we divide by the new volume of 0.8 Now that we have reached 80 mL total volume and we get a hydro knee, um, concentration of 0.50 And when we take the negative log of that, we get a pH of 1.301 at 80 mL. We recognize this is the equivalence point. This is a strong acid, strong based hydration. Therefore, at the equivalence point, the pH is seven. No calculation is required at 100 mL of Thai trahant Being added, we are now post equivalents. So what is in excess in solution is now the strong based potassium hydroxide, so to solve. For Ph, the first thing we need to do is solve for the hydroxide concentration in excess that will be equal to the volume of potassium hydroxide in excess. If we've added 100 mil leaders and the equivalence point volume was 80 then we have 20 mil leaders in excess that corresponds to 200.20 leaders. We multiply that by the molar ity of potassium hydroxide to give us the moles of hydroxide in excess. And then we divide by the total volume, which is now 0.140 leaders, the 40 mL plus the 100 mL, and we'll get a hydroxide concentration of 1.429 times 10 the negative, too. To get pH, we first need to convert the hydroxide into hydro knee, um, by dividing it into K W. And then take the negative log of the 7.0 times 10 the negative 13 to get our ph of 12.155

Discussion says, for asks how many milliliters of a 0.15 Moeller solution of potassium hydroxide will be required to titrate 40 mills of a 400.656 Miller solution of each three people. Four. And gives his bounds chemical equation that I've written here. So it started with what we know about the acid tells us that we have 40 mills. We'll write that on leaders because that's the units that polarity is in. Is your four leaders of H three appeal for And it tells us the concentration is point 06 56 Moeller, which is points your six by six moles of the acid over one leader each. Three people for that would convert between moles age three people for and passing hydroxide himself in bounds. Chemical equation You see that one mole each three people for combines with two moles of potassium hydroxide And so we should be able to cancel unit's leaders each. Three people are moles h three people for like with moles of college. We do this. We would get 30.0 zero 52 moles of college. Then we can figure out what volume of the port 15 Miller solution will get us that money moles. Let's write this out. We have 0.15 more solution. Okay? Which and that would equal the number of moles. We just calculated that we need 1005 to goals of Ko age and some volume of that are rated as acts. Some number of leaders of K O H. That is one type of will get us that number of bulls. So say it again. Uh, this is number of moles we need and we can calculate the volume of the stock solution were given to figure out how many leaders we need to reach that number of moles. So if you re arrange to get X by itself, you would find the number of leaders. If you convert to the leaders the most. By my 1000 is 34.99 No leaders ko each passing hydroxide


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