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12108 Hd210 20 30 40 mL NaOH50...

Question

12108 Hd210 20 30 40 mL NaOH50

12 10 8 Hd 2 10 20 30 40 mL NaOH 50



Answers

How much NaOH are in 250 $\mathrm{mL}$ of a 3.0 $\mathrm{M}$ NaOH solution?

Hello Today we'll be talking about Chapter 14 Question 22 which gives us a sodium hydroxide solution and a wage in water and asks us to calculate how much anyway. H Ingrams is in the solution. So what's look at our givens? First, we're told that we have a three Moeller any wage solution. And of that we have 250 milliliters. Now we know that three Moeller because mole equals moles of saw Ute heard of leaders of solvent. So we know that that this is the same as 3.0 Mol of any a wage over one leader of solution. We can also convert 250 milliliters in two leaders by multiplying it by one leader over 1000 Villiers, which is equal to 0.250 leaders. And so now we have our volume of solution in leaders. We have our concentration in moles per leader, and it looks like we can cancel out our leaders if we multiplied together. So let's multiply No 3.0 moles of any a wage in one leader times 0.250 Leaders of solution Our leaders can cancel out and we're left with the right number of moles in the solution, which is going to be 0.75 Moles does any wage. Now we need to get this number into grams. And so we need the molecular weight of sodium hydroxide. And we can find this by just adding up the constituent elements the molecular weight of sodium oxygen and hydrogen using any standard periodic table. And we can get a molecular weight of 39 0.997 grams of any wage. Permal for one mole of any Ohh! And so if we take this number now and we multiply our grams for mole by the number of moles of any wage, we should once again be able to cancel the units and be left with grams. So if we take 0.75 moles was n a h times 39.997 grams of any ohh divided by one mole of N a. O. H. We can see that are moles. Cancel a topping above a blow and above the line, and we're left with grams of any wage. And so our final answer is going to be 30 grams of any wage which we can rewrite and scientific notation as 3.0 times 10 to the first power grams of n a. O. H. And we want to limit ourselves to two significant figures because we have two significant figures limiting us in our concentration and in our volume. Because two and five are the only two significant figures in this number, only zeros that come after a decimal point and after another number are significant. So our final answer is 3.0 times 10 to the first grams of any a wage in this given solution.

If we're trying to reach a target ph of 4.7 and we have a solution that contains hydrogen oxide late and we're adding sodium hydroxide. We need to convert some of the hydrogen. Um, well, I guess the hydrogen aqsa late to just pure aqsa late to see 204 tu minus. So that means will be working with the PK A to because we will have stripped off all of the hydrogen is and will have some concentration of just Oxlade present. So our target ph is 4.7. We're gonna use K two. So the negative log of K two in order to get p K two will still have a buffer solution. So we're using the Henderson hassle Baulch equation and the moles of the base. The base being the completely, uh, um deep throat nated form of oxalic acid or just the Oxlade itself. Um, we need to because we're just starting with h c 204 minus than when we add the any o. H. We're making the Aqsa late so we don't have any aqsa late to begin with. And every mole of sodium hydroxide that we add will make a mole of the ox late. So we have an unknown amount of sodium hydroxide that we need to add, multiplied by the concentration of sodium hydroxide to get the mold sodium hydroxide, which will then correspond to the moles of Oxlade formed. But we already have some H c 204 minus, which we can calculate by taking the volume and multiplied by the molar ity. And for every mole of sodium hydroxide we add, we will decrease the moles of the H C 204 minus by that same amount. So the rest is just algae breath will do some rearrangement. Collect are like terms after collecting are like terms. Well, then solve for X. So the first step I did here was I simply took the negative log of this and then moved it over and then took the anti laudable sides. And I got this that I multiplied both sides by this denominator down here and 10 to the 100.56 is 3.21 than the Multiply it by the denominator will still have this over here is our numerator and we get this equal 2.12 times are 0.1 to 0 x. Well, then collect are like terms. I'll, um, ad 0.385 x double sides and then divide herbal sides by 0.505 may get 0.635 leaders or converted to mill leaders is 63.5 milliliters.

So we use a delusion equation here on this one. Also, if you want to make 500 mil, leaders have a point to Moller. Solution will take an unknown volume of the stock solution that is 15 Moeller. The unknown volume, when calculated, is 6.67 mil leaders. So I would take 6.67 mL of the 15 Mueller solution they looted to 500 mL and I would have a point to Mueller solution.

When calculating polarity of sulfuric acid by knowing smalls and dividing it by its leaders. So polarity sulfuric acid is simply most sulfuric acid divided by leaders. We can calculate the most sulfuric acid by taking the volume of sodium hydroxide that was required to react with sulfuric acid, converting its leaders and then converting the volume, or the leaders of sodium hydroxide to moulds, sodium hydroxide using the polarity of sodium hydroxide. Then we recognize the story. Geometry of the reaction has given to one to mull sodium hydroxide required for everyone. Mold that sulfuric acid that then gives a small sulfuric acid in our numerator and the denominator. The volume that was tight traded was 20 milliliters. Sulfuric acid will convert that to leaders by dividing by 1000 that then gives us similarity of 0.234 molar sulfuric acid


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