find the pK for the following ionization constants, K=
1.8*10^-5 and 1.1*10^.2
Can't you do these?
pKa = -log(Ka). Just punch in the numbers and out comes the answer.
Now I can. Thanks. Our labe manual is an "experiment" for next year's classes, so a lot of info has been left out in error and we have to fend for our selves. I really appreciate the help.
To find the pK for ionization constants, we can use the formula:
pK = -log10(K)
Let's find the pK for the given ionization constants:
For K = 1.8 * 10^-5:
pK = -log10(1.8 * 10^-5)
pK = -log10(1.8) - log10(10^-5)
pK = -log10(1.8) - (-5)
pK = -log10(1.8) + 5
Using a calculator to find the logarithm of 1.8:
pK ≈ -0.2553 + 5
pK ≈ 4.7447
For K = 1.1 * 10^0.2:
pK = -log10(1.1 * 10^0.2)
pK = -log10(1.1) - log10(10^0.2)
pK = -log10(1.1) - (0.2)
Using a calculator to find the logarithm of 1.1:
pK ≈ -0.0414 - 0.2
pK ≈ -0.2414
Therefore, the pK values for the given ionization constants are approximately 4.7447 and -0.2414, respectively.
To find the pK for each ionization constant, first, let's recall the definition of pK. The pK is the negative logarithm (base 10) of the ionization constant (K). In other words, pK = -log(K).
1. For the first ionization constant, K = 1.8 * 10^(-5). To find pK, we take the negative logarithm of K:
pK = -log(1.8 * 10^(-5))
Now, to evaluate this expression, we can use logarithm rules. Since the base is 10, we have:
pK = -log(1.8) - log(10^(-5))
The first part, -log(1.8), can be calculated using a scientific calculator or logarithm tables. For simplicity, let's assume it is approximately equal to 0.255. The second part, log(10^(-5)), equals -5 because the logarithm of 10 raised to any power is equal to that power.
Therefore, we have:
pK ≈ 0.255 - 5
pK ≈ -4.745 (rounded to three decimal places)
So, the pK for the first ionization constant is approximately -4.745.
2. For the second ionization constant, K = 1.1 * 10^(0.2). Again, we apply the same steps:
pK = -log(1.1 * 10^(0.2))
Breaking it down:
pK = -log(1.1) - log(10^(0.2))
The first part, -log(1.1), can be determined as before. Let's assume it is approximately equal to 0.041. The second part, log(10^(0.2)), equals 0.2 since the logarithm of 10 raised to any power is equal to that power.
Therefore, we have:
pK ≈ 0.041 + 0.2
pK ≈ 0.241 (rounded to three decimal places)
So, the pK for the second ionization constant is approximately 0.241.