pH of Strong Acids and Bases - Video Tutorials & Practice Problems
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concept
Strong Acids and Bases
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Hey, guys, In this new video, we're gonna take a look at the connection between Ph and P. O. H and how they relate to strong acids and strong bases. So remember, we've talked about this before. We're gonna say that strong acids and strong bases are considered to be strong electrolytes. Remember strong electrolytes, Completely ionized. That means they break up 100%. You may even hear that these strong acids and bases air called highly electrolytic. That also means they break up completely highly electro politic. So if we take a look here, we have HCL in N a o h. Remember, from the rules we've learned, we know HCL is a strong binary acid, and we know n a O. H is a strong base because we have a group one ai on and a positive connected to O. H minus. So what happens here? We have a single arrow going forward because they break up 100% to give us thes ions. We'd say that the product side is highly favored and just realize if we have a strong acid or a strong base, I've said it before. We don't have to use a nice chart in order to find pH or P O. H. So if you have a strong acid, take the negative log to find pH. If you have a strong base, take the negative log to find P O. H. So let's take a look. At example, one.
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example
pH of Strong Acids and Bases Example
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so an example? One, it says calculate the pH off a 0.782 Moeller solution off. Calcium hydride. No, we know that calcium hydride is a strong base. Because remember, we have calcium, which is C A two plus connecting with h minus are hydride ions. So what happens here? The two from here came here. The one from here came here. That's how we got calcium hydride. And what we should realize here is that there are two calcium hydride in the formula. Now, this is what you need to remember. When it comes to strong basis for strong bases. You have to be very careful. We're going to say that this is not the real concentration of our strong base. You're gonna say when you have strong bases, you have to look at the number off O H minus H minus and H two minus 02 minus. So you have to look to see how many of each you have in order to determine the rial concentration off that strong base. In this particular question, we have to h minus is and because we have two of them, we have to multiply this by two that will give us the correct concentration off our strong base. Okay, so you have to remember that. So, for example, if I had C a o. H. Two and there was a 0.100 Moeller of it again because you have 20 H is you'd have to multiply that by two. If you have calcium of mind, you would have to an h two. So again, you have to multiply the concentration times to to get the correct concentration. Now, what we're gonna do is we're gonna multiply the initial concentration times to to get the rial concentration. Okay, When we do that, we're gonna get our correct concentration. That's the number that we're gonna plug in. So when we do that, and I'm gonna remove myself guys from the image so we have more room to work with When we do that, we're gonna get a new number a new value for r O. H. Minus concentration. So here, we're gonna get 0.1564 Mueller, that becomes our new concentration. So we're going to say p o. H equals negative. Log off O H minus. So it becomes negative. Log off 0.1564 Moeller. Now, some of you may be asking, but Jules, this is saying the concentration of O. H minus. We don't have O. H minus. We have h minus. How do they relate? Just realized that all four of these ions can be treated as being the same. They all feed into the concentration of O H minus. Okay, so h minus. Just understood as a general term for the strong basic ion off a strong base. So if I have to h 20 h minus is that's equivalent to two h minus is equivalent to two NH two minuses. Same thing. So when we take the negative log of this, this will give us our p o h. It's gonna give us 0.81 for our p o h. But remember, let's pay attention. I didn't ask for the pH I asked for. I didn't ask for the p o. H. I asked for the pH. So we need to use the next equation. PH plus p o h equals 14. Plug in what we know. For P o h. 0.81 we need ph. So subtract 0.81 from both sides. So ph here equals 13.19 So that will be our pH for this first question. Hopefully, guys didn't get lost. But what I was saying Just remember for strong bases you have to look to see how Maney o H minus is You have h minus is you have an H two minuses. You have an 02 businesses you have their number, determines the correct concentration of our strong base. And right now, what you need to realize is we only do this for strong base. At this point. If we had a strong acid such as H two s, 04 that strong acid has too acidic. H is. But we wouldn't need to multiply the concentration times too, because when it comes to acids, the first age coming off extremely strong. But the second one that comes off is extremely weak. And because the second H plus is extremely weak, we wouldn't need to multiply the concentration times too. So we only do this multiplication of concentrations when we have strong basis. Later on, when we deal with acid based titrate, shins and buffers, we'll talk about how assets play a role
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Problem
Problem
An aqueous solution of HBrO4 has a pH of 4.34. Find the molar concentration of HBrO4 solution.
A
4.6 × 10−19 M
B
2.6 × 10−6 M
C
2.2 × 10−10 M
D
4.6 × 10−5 M
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Problem
Problem
Calculate the pH of a 25 mL of 5.45 × 10−2 M LiOH solution.
A
1.264
B
12.736
C
0.338
D
11.134
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Problem
Problem
HI is a strong acid (Ka = 3.2 × 109). Calculate [H+], [OH−], pH and pOH of a 7.1 × 10−2 M HI solution.
A
[H+] = 7.1 × 10−2 M
[OH−] = 1.4 × 10−13 M
pH = 1.15
pOH = 12.85
B
[H+] = 7.1 × 10−2 M
[OH−] = 1.4 × 10−13 M
pH = 8.50
pOH = 5.50
C
[H+] = 7.1 × 10−2 M
[OH−] = 7.1 × 10−16 M
pH = 1.15
pOH = 15.15
D
[H+] = 7.1 × 10−2 M
[OH−] = 7.1 × 10−12 M
pH = 1.15
pOH = 11.15
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