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50.0 mL sample of# 6.0 M solution of HCI diluted 200, mL What IS the ncw conceniration? 4)24.0 M B) 6.0 M 0)2i0M DJ 2.00 M E) 150...

Question

50.0 mL sample of# 6.0 M solution of HCI diluted 200, mL What IS the ncw conceniration? 4)24.0 M B) 6.0 M 0)2i0M DJ 2.00 M E) 150

50.0 mL sample of# 6.0 M solution of HCI diluted 200, mL What IS the ncw conceniration? 4)24.0 M B) 6.0 M 0)2i0M DJ 2.00 M E) 150



Answers

If you dilute 20.0 $\mathrm{mL}$ of a 3.5$M$ solution to make
100.0 $\mathrm{mL}$ of solution, what is the molarity of the dilute
solution?

Question here wants us to calculate the final volume using the equation. See, one of you one is able to see two v two. So the answer that you'll get is going to be for eight year 80 leaders from be here 250 milliliters. The 1st 1 should be milliliters. Rather first, see here you're gonna get 600 middle leaders and for D here, you're going to get 180 milliliters.

This problem is multiple steps just like 57. It's a strong base now, being tight, traded with a strong acid instead of a strong acid being tight trade with a strong base. But still the first thing that we need to do is determined the equivalence point volume. To do that, we'll take the volume of barium hydroxide 80 million leaders that is being titrate ID converted to leaders. We get eight leaders, then convert the leaders barium hydroxide, two moles, barium hydroxide using the concentration of 20.1 moles, barium hydroxide per leader. Then we'll go from moles. Barium hydroxide, two moles of the hydrochloric acid needed to neutralize it. It being to toe one because barium hydroxide has to hydroxide while three HCL only has one hydrogen ion. Then when we have moles HCL we can convert moles hcl into leaders hcl using the polarity HCL and then from leaders hcl to mill leaders. And we see the equivalence point volume being 40 mL of hcl. So when we start out at zero mill leaders, the hydroxide concentration is going to be equal to two times the barium hydroxide concentration because there's two hydroxide for everyone Barium hydroxide. So we have a point to moller hydroxide concentration when we start without having added any HCL. The hydroxide concentration, then, is 0.2 Moeller. To get pH, we take the negative log of the hydro knee Um, concentration, which is the hydroxide concentration divided into K. W. And we get a pH of 3.301 Then a 20 million leaders were halfway to the equivalence point. The hydroxide concentration is going to be the volume of barium hydroxide that we're starting with, multiplied by the hydroxide concentration originally in that barium hydroxide solution of point to Moller, which we established up here and then subtract off the moles that react, which will correspond to the moles of strong acid, added the most of strong acid added will be the 20 mil leaders expressed his leaders point to multiplied by its concentration of 200.0.4. Then we divide that by the new volume, which is 80 mL 800.8 liters plus 20 mL, two leaders and we get a hydroxide concentration of 20.80 pH then is going to be equal to the negative log of the hydro knee um concentration, which is the hydroxide concentration divided into K. W. And we get a pH of 12.903 at 30 mil. Leaders were still pre equivalent, so we still have excess hydroxide hydroxide. Concentration, as we did appear, will be equal to the moles we start with, which is the volume times the mole arat e minus the moles that react, which will be the volume of HCL added, multiplied by its mill arat e divided by the total volume which is the 0.8 plus the 30 mL point of three leaders that we added, and we get a p a a hydroxide concentration 2.3636 to get the pH will take the negative log of the hydro knee, um, concentration, which again, is the hydroxide concentration divided into K. W. And we get a pH of 12.561 at 40 mil. Leaders. We established this with the equivalence point. So for a strong acid, strong based hi Trish in no calculation is required and the pH is always seven at 80 mil. Leaders were now post equivalents, so we have excess hcl that has been added. So we're going to have a excess hydro knee. Um, concentration, where in 80 mL equivalents point was 40 mL. So we're 40 mL past the equivalence point, so we have an excess of 40 million liters of HCL that has been added 40 mL at a concentration of 400.4. When multiplied together, will give us the moles of hydrogen IAM in excess to get concentration, will then divide the moles of hydrogen IAM by the total volume, which is the 80 million liters plus the 80 mL added or 800.16 leaders, and we get a hydro knee, um, concentration of 0.1, which corresponds to a pH of one.

Here 25 millimeter up 1.50 Mueller. A cell is diluted to 500 millimeter. We have to calculate the molar concentration of the diluted A. C. Now during dilution number of moles of solute remains unchanged, but the concentration changes so we can write this relationship more clarity of the concentrated solution times the volume of the concentrated solution will be equal to the modularity of the diluted solution times the volume of diluted solution. Normal clarity of the consultative solution is 1.50 Times volume in later will be 25 by 1000 is more clarity of the dilute solution times the volume 500 millimeter means 500 by 1000. Later this 1,000,000 will begin Salau. There are four more clarity of the diluted solution will be 1.50 times 25 Divided by 500. So the more clarity Of the diluted solution will be 0.075 morning.

We were also a relatively simple delusion. Calculations. We use the delusion equation the original volume times the original concentration will set that equal to the new volume in the new concentration. The new volume is what we don't know. So in solving for the new volume, we will get 100 milliliters in order to go from 10% to 2.5% mass volume HCL the next one If we have 25 milliliters at five Moeller to get one Moeller we need to dilute it to 125 milliliters and then if we have 25 milliliters at six Mohler we need to dilute it to 300 milliliters in order to get 3000.5 Mueller hcl.


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