As your post title said, you have to use the Henderson-Hasselbalch equation, which is: pH = pKa + log ([A-]/[HA]).
Now since there is no calculator for the science section of the DAT, we have to estimate numbers with a degree of error. Our target is pH buffer of 4. So, we can begin by plugging in for pKa. The pKa is -log(Ka), or -log (4.5 x 10^-4). The pKa is estimated to be around 3.2 or 3.3. To estimate this, simply take the value of the exponent (4 in this case because it's 10^4), and then compare it to the base, or 4.5. If the base is greater than 1, then the estimated value will be less then the value of the exponent. In this case, since 4.5 is greater than 1, then the estimated answer will be less than 4, or about 3.3.
Now we can apply the same rules to the second part of the equation, log ([A-]/[HA]). The value for [A-] comes from NaNO2, because it is soluble and produces NO2 ions. Similarly, [HA] comes from HNO2 concentrations.
So, we need to come up with a ratio of [A-] to [HA] that gives us a value of less than 1. We know it has to be less than 1 because pKa is 3.3. Using algebra, where 4 = 3.3 + ratio, the ratio has to be about 0.7. Now, for the log of the ratio to be less than 1 (because our target is 0.7), we have to get a value greater than 1. Simply put, log (big ratio) = small output number. In our case, we want our answer to be a small number, so the log of the ratio has to be a big number, or [A-]/[HA] has to be bigger than 1.
Using this information, where log (big ratio) = small number, then we can eliminate answer choices A and D immediately, as they give ratios less than 1 which doesn't satisfy our "big number" criteria. We are now left with B and C. The ratio for B is about 1, and the log of 1 is 0. This doesn't work because 4 does not equal 3.3 + 0. So we are left with answer choice C.
To check, we know that log (0.50/0.11) is approximately the same as log (5), because (0.5/0.11) is approximately 5. Now, using logarithmic power rules, we rewrite the expression log (5) = x, where we want to solve for x, as 10^x = 5. By simple math, we know that x has to be between 0 and 1 to give us an answer of 5 for the second equation, since 10^0 = 1 and 10^1 = 10. This gives us 0<x<1, which matches our predicted answer of 0.7.
This is a detailed explanation of how to go about the problem including all of the necessary math and estimations. (Notice how we never even needed a calculator!) Once you practice more and more of these types of problems, the estimations of logs will become second nature. Good luck!
