If a buffer system containing hydrofluoric acid and fluoride (pka = 3.17) initially has a ph of 3.22 and base is added until a new ph of 4.69 is obtained, how has the ratio of fluoride to hydrofluoric acid changed?

A. from 1.12 : 1 to 33.1 : 1

Explanation:

#Plato

Use the Henderson-Hasselbalch Equation:

Plug in values given (let's start with the initial pH of 3.22)

3.22 = 3.17 + log(x) | Since we don't know the concentrations, make it x.

Subtract 3.17 from 3.22 (we're finding x)

.05 = log(x)

Take the inverse of log to find x

x = 1.12

Now, "if x is less than 1.0, then there is more acid than base; if greater than 1.0, there is more base than acid." So we have more base than acid so the initial ratio is 1.12:1.

To find the new pH, plug in values into the equation again:

4.69 = 3.17 + log(x)

You'll find that x = 33.1. Once again, x is greater than 1 so there is more base than acid. Plus, you can see that the solution's pH went up from 3.22 to 4.69 so it got more basic.

Thus, the ratio went from 1.12 : 1 to 33.1 : 1

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)