some general chem questions
Started by mass
Hey, the fact, the Ka's are given, makes life so much easier. Ka is a measure of the acidity content of a compound. The higher the Ka, the more acidic the substance. Since you're ranking in increasing order, start from the compound with the lowest Ka value, & work your way up to the highest. When the smoke clears, we get: 2,, 3, & 1 in that order. Hope this helps & good luck studying.
- Donjuan
- Donjuan
The key here is that the more negative the exponent, the less acidic and the more closer the exponent is to being positive, the more acidic it is, since a larger Ka value means the compound is more acidic. We're comparing orders of magnitude, so only focus on the exponents, unless atleast two or more were the same value. Hope this helps and let me know if you need anymore help!
_ Donjuan
The scientific notation exponent actually gives you a rough estimate of the pH, and thus the acidity (the lower the pH, the higher the acidity, obviously)
pH = -log[H+] = -log(Ka)
1. HF- KA 7.0*10-4 = -log(7x10^-4) = 4.84 ---- Note: since the number 7 is closer to 10 than 1, the pH is correspondingly closer to 5 than to 4; if it was 1x10^-4, the pH would simply be 4
2. CH3COOH- KA 1.7*10-5 = -log(1.7x10-5) = 5.23 ---- same as #1, the number is slightly bigger than 1, so the pH is slightly bigger than 5
3. HS04- KA 1*10-2 = -log(1x10^-2) = 2
pH = -log[H+] = -log(Ka)
1. HF- KA 7.0*10-4 = -log(7x10^-4) = 4.84 ---- Note: since the number 7 is closer to 10 than 1, the pH is correspondingly closer to 5 than to 4; if it was 1x10^-4, the pH would simply be 4
2. CH3COOH- KA 1.7*10-5 = -log(1.7x10-5) = 5.23 ---- same as #1, the number is slightly bigger than 1, so the pH is slightly bigger than 5
3. HS04- KA 1*10-2 = -log(1x10^-2) = 2
i got the order to be 3,1,2 because I used pka values. H2s04 has the lowest pka value so its the most strongest acid no?
secondly I have a few more questions:
calculate the theoretical ph of a mixture of 20ML 0.5M NH3 and 20ml of 0.1M HCL (PKA OF NH3=9.2)
secondly I have a few more questions:
calculate the theoretical ph of a mixture of 20ML 0.5M NH3 and 20ml of 0.1M HCL (PKA OF NH3=9.2)
Hello, i'll apologize right away for the french terms that i might use, because i study in french.
But basically, your reaction gives you: NH3+HCl --> NH4+ + Cl-
Since HCl is a strong acid, it'll fully react with your base, NH3.
Therefore 0,1M of NH3 will react with 0,5 of NH3 and will give you: (0,5M-0,1M=)0,4M of NH3. And what has reacted has to give a product, and therefore give you (0M+0,1M=)0,1M of NH4+ and Cl- from the 0,1M of acid (HCl) that reacted.
With those information you can find the pH:
First, they give you the pKa for NH3, you change it into pKb (14-9.2=)4.8
Then, you use the hendersohn-hasselbach equation using the information you found earlier: pOH= pKb+log (nh4+/nh3) = 4.3+log(0.1/0.4) = 4.19
With the pOH you can now find the pH: 14-4.19= 9.81 (the answer!!!)
Hope i did the right thing, can anyone back me up on that?
But basically, your reaction gives you: NH3+HCl --> NH4+ + Cl-
Since HCl is a strong acid, it'll fully react with your base, NH3.
Therefore 0,1M of NH3 will react with 0,5 of NH3 and will give you: (0,5M-0,1M=)0,4M of NH3. And what has reacted has to give a product, and therefore give you (0M+0,1M=)0,1M of NH4+ and Cl- from the 0,1M of acid (HCl) that reacted.
With those information you can find the pH:
First, they give you the pKa for NH3, you change it into pKb (14-9.2=)4.8
Then, you use the hendersohn-hasselbach equation using the information you found earlier: pOH= pKb+log (nh4+/nh3) = 4.3+log(0.1/0.4) = 4.19
With the pOH you can now find the pH: 14-4.19= 9.81 (the answer!!!)
Hope i did the right thing, can anyone back me up on that?
