Acid-Base Equivalents

here we can say that equivalents are used to measure the number of H plus and O. H minus ions in acids and bases respectively. Were going to say when it comes to the equivalent of an acid it's the amount of acid that contributes one mole of acidic H plus ions and equivalent of a base is the amount of bait that contributes one mole of O H minus ions. For example we have one bowl of hydrochloric acid it possesses in it. One H plus ions. That one H plus ion is equal to its one equivalent. Here we have one mole of calcium hydroxide, it has two hydroxide in it. So that's two equivalents of calcium hydroxide. Now we'll go into further detail where what if the moles of the acid and base are different That can help vary the number of equivalents. So click on the next video let's continue talking about um equivalent when it comes to acids and bases.

Now to calculate the number of equivalents of acid or base, we simply multiply N by the number of moles of acid or base and equals moles of H plus or oh H minus. Now, with this whole concept we also have million equivalents. It's the common unit used to express equivalents, remember that one equivalent is equal to 1000 million equivalents. So if we're talking about acid equivalents, remember an equivalent is equal to N. Which is just the number of H plus ions times the moles of the acid. And for base it's equivalent is equal to end, which is the number of which minus ions times the moles of the base. Keep this in mind when looking at the equivalent for either acids or bases.

calculate the number of equivalents in each of the following. So for the first one we have here is one mole of phosphoric acid. So this is an acid and remember it is N. Which is the number of H plus ions for the acid times the moles of the acid here. Plus forecast it has in it three H plus ions. So and equals three times. Were going to say here one mole. So that equal three equivalents for phosphoric acid. Now let's look at the base. Here, we have our B O. H. Which is a base. So this would be the equivalent is equal to end, which is the moles number of O H minus times. It's moles. We know how many O H minus we have in this compound. We just have just one. So, and is one. What we have to do now is figure out the moles of this particular base. So we're gonna have to convert grams to moles. Take into the account we have one RB, one oxygen, one hydrogen. So that's one mole. Okay, Looking up their masses on the periodic table and adding them together gives me a molar mass of 102. g grams cancel out. And now we're gonna have .0- moles of this space. So that number goes here. So one times that number means I have .0-6 equivalents of my base. So this is how you identify and calculate the equivalence of either an acid or a base.

So the equivalent weight represents the mass in grams of one acid or base equivalent. Here, we're gonna say equivalent weight equals the molar mass of the acid or base, divided by N. For an acid, and equals the number of H plus ions. For base, it equals the number of O. H. Minus ions.

Here we need to calculate the equivalent weight of sulfuric acid. So, remember, equivalent weight equals the more mass of the acid. In this case divided by N. Which is it's total number of H plus ions. So, so if uric acid has in it to H plus ions, so, and will be too, and then molar masses, just calculating the combined mass of the two hydrogen, the one sulfur and the four oxygen's taken from the periodic table. If you do that correctly, you'd have 98 0.86 g as the molar mass grams per mole as the molar mass for sulfuric acid. So that number gets divided by two, so it's 49.43 g per mole as the equivalent weight of sulfuric acid.

The concentration of acid or base in equal solutions is represented by normality. Now, normally itself represents the number of equivalence per liter of solution and recall that molar itty itself equals moles over leaders, moles of solute over leaders of solution. Here, we can use this information to apply to normality. Non normality itself, we said can equal equivalent per leaders of solution or we can say that normality equals n. Remember. And for acids is the number of H plus ions and for bases is the number of O H minus ions times your polarity, which is capital. M. So remember. These are the two formulas we can use to establish relationships with normality of solutions.

Calculate the normality of each of the following solutions. So here for a we have 4.6 times 10 to the negative. Two more of sodium hydroxide. Remember when they give us more clarity. Normality is just simply end times polarity. And here would be the number of O. H minus ions for the base here there's only 10 H. Within sodium hydroxide. So which is simply be one times 4.6 times 10th and negative two. So my normality will be just 46 times 10 to the negative too. For be a little bit more is required For B. We have .35g of phosphoric acid in one leader. So this is telling me and I'm using the equation normality equals equivalents divided by leaders of solution. So here would be one leader on the bottom. And what we need to do is figure out the equivalence of this base. Remember for based equivalents are and which is the number of H plus times. It's moles eight street P. 04 has in it. Three H plus ions. So maybe three times now. We need the moles of phosphoric acid. So .35 g of phosphoric acid. We're gonna say for every one mole of phosphoric acid, what are the gramps When we had the three hydrogen is the one phosphorus and the four oxygen's we get back a mass of 97.994. So here that's gonna give me my moles, Which comes out 2. moles. So plug it over here. So that's going to give me when I multiply. It's gonna give me .0107148 equivalents plug in here. It's getting divided by one. Should be the same exact number. .35 has 266. So this would just be .011 normality. So, there will be my answers for options A and B. Here

Calculate mass (grams) needed for the following base equivalent:0.18 mEq of Mg(OH)₂.

• A. 5.3×10^–3 g
• B. 0.010 g
• C. 10 g
• D. 5.3 g

Identify the acid that possesses an equivalent weight of 63 grams.

• A. H₂C₂O₄
• B. HCl
• C. HNO₃
• D. H₂CO₃

Determine volume (mL) needed to prepare a 0.73 g of Ca(OH)₂ solution with 1.25 N.

• A. 7.9 mL
• B. 20 mL
• C. 1.6 mL
• D. 16 mL