# Tutor profile: Albert L.

## Questions

### Subject: Chemistry

A 50.0 mL, 0.5M solution of hydrochloric acid is mixed with 20 mL of a solution of potassium hydroxide of unknown concentration. A test is later done, and the pH of the resulting solution is 5. What is the concentration of the solution of potassium hydroxide? Assume all of the ions are fully ionized.

Knowing how acid-base reactions work to answer this problem is critical! Given that pH is defined as the inverse logarithm of the concentration of hydrogen ions, -log[H+], this will help us approach our answer. We can first calculate the amount of hydrogen ions in the initial solution, and then find out how much was neutralized. 50.0 mL * 0.1M of HCl means 50 mmol of H+. With the 20 mL of extra solution, normally the initial pH would mean 5 mmol/70 mL of solution, which is 0.07143 M. This gives us a pH of 1.146. Given that the ending pH is 5, which doing the reverse calculation, results in a concentration of 1*10^-5 M, that means the concentration decreases by 0.07142 M. The OH- ions of the KOH solution has to match that concentration. So given we known that the concentration of OH- added is 0.07142 M, we can find the solution concentration of the original solution. 0.07142 M * 70 mL = 20 mL * x M. This gives us the original concentration of 0.2499 M.

### Subject: Calculus

Find the derivative of the function: 5+6x^2+e^cos(x) = f(x)

There's two main things to consider: power rule, and chain rule. Let's do the simple part first. We can find the derivative of 6x^2+5 with the power rule, which allows us to find the result of this portion as 12x. e^cos(x) uses the chain rule, which is a bit more difficult. Chain rule basically means that the outer function we take the derivative of last. So the derivative of cos(x) is -sin(x). Then we take the outer derivative, which is in the form e^x. e^x derivative is e^x. So we multiply by the inner derivative: -sin(x)e^cos(x) and then combine with the previous part to get our answer: -sin(x)e^cos(x) + 12x.

### Subject: Algebra

Find the solution to the system of equations: x+3 = y; 2y-x = 5

Thankfully, x+3 is already defined as y, so substitution (plugging in y in terms of x) is going to be the easiest way for this to be solved! 2(x+3)-x = 5 gives us 2x+6-x = 5, which means that x+6=5, and so x = -1. With that, we can easily substitute x back into the first equation to find that y = 2.