Let's begin part A by considering what we have. We have a solution of a weak acid, butanoic acid. Now, I know it's a weak acid for a couple of reasons. It has the COO group, that's an organic acid group, and they're all weak. A Ka value is given, that certainly is an indication that's a weak acid. And we haven't added any of the sodium hydroxide, which is a strong base at this point. So all we're going to do at this point is consider a weak acid in water. Now, I could write the whole formula, or I could just let HA represent my weak acid. Because it's kind of cumbersome to write this whole thing out. So I'm going to let HA represent the weak acid. This H is actually the donatable proton which is sitting right here. It's in the presence of water. And we end up with that proton swap, the H+ from the acid is donated to water. That'll leave me with A- and H3O+. Now you might have chosen to write this reaction HA in equilibrium with H+ and A-. That is okay as well, but I always include the water when I'm doing a weak acid or a weak base problem. My table would be an ICE table because it's an equilibrium. Initially, there is a concentration of butanoic gas listed right here of 0.249. I don't care about the water. There is no A- and there is essentially no H3O+. We know that water has a tiny amount present. We will subtract x, we will add x. And we will be left with 0.249 minus x, x, and x. Since this is an equilibrium of a weak acid, we know that Ka is equal to the A- concentration times the H3O+ concentration, divided by the HA. That doesn't look like an H, HA concentration. And we can plug the values in that we know. Ka is given up there as 1.54 times 10 to the -5. And that is going to be equal to x times x, or x squared, divided by 0.249 minus x. Because Ka is small, at this point we can ignore this x term or at least try to, and multiply both sides by 0.249 to get it off the denominator. This will give me 1.54 times 10 to the -5, times 0.249, equal to x squared. When I multiply this two values, and also take the square root, I will obtain a value for x of 1.96 times 10 to the -3. And this is a very small number compared to that, so it is a legitimate assumption that we made. You could take this value divided by this value, okay. This number here divided by this number, and you would see that it is less than 5% of that value. Now x represented, if we look in our table, H3O+ concentration. So we are now able to obtain the pH, pH which we know is the negative log of the H3O+ concentration. I'll plug in 1.96 times 10 to the -3 for that H3O+ concentration. This will give me a value of 2.708. Now, let's stop and think about the reasonableness of this value, 2.708. Our pH is less than seven. It is a acid, so we'd expect it to be less than 7. It's not extremely low, a strong acid would be very, very low if it had a 0.249 molar concentration. So we'd expect it to be higher than 1, but not higher than 7