posted by tt .
1. A solution of sodium oxalate has a pH of 7.82. The [OH-] in mol/L must be which of the following:
a) 6.18 b) 1.5 x 10-8 c) 6.6 x 10-7 d) 7.82 e) -7.82
2. When 0.93 mol of O2 and 0.56 mol of NH3 are mixed together and allowed to come to equilibrium according to the equation:
4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g)
At equilibrium it is found that there are 0.77 mol of O2. Which substance has a change in number of mols of 0.192 as the system comes to equilibrium?
a) H2O b) NH3 c) O2 d) NO e) no substances change by 0.192 mol
3. At normal body temperature, Kw has a value of 2.42 x 10-14. What is the pH of a neutral solution at this temperature?
a) 7.2 b) 13.6 c) 6.8 d) 7.0 e) 6.2
You would do well to make three separate posts. It takes too long to work through three problems; therefore, many profs will let the post go if they can't do it all.
a. pH + pOH = pKw = 14
Use the above equation to calculate pOH, then pOH = -log(OH^-).
b. You must learn to incorporate arrows. Without arrows you can't tell the difference between products and reactants.
I would look at this.
mols O2 initially = 0.93
moles O2 @ equil = 0.77
moles O2 used = 0.16
Now convert 0.16 moles to NH3, H2O, and NO and you will find the answer. You convert by using the coefficients in the balanced equation. For example,
moles NH3 = 0.16 mol O2 x (4 moles NH3/5 moles O2) = 0.16*(4/5) = 0.128 so NH3 can't be it.
moles NO = 0.16mol O2 x (4 moles NO/5 moles O2) = 0.16*(4/5) = 0.128. It was zero initially so NO can't be it.
Continue until you find the right one.
H2O ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 2.42E-14
Remember (H^+) = (OH^-)
Substitute and solve.
Don't be surprised when the answer comes out something OTHER THAN 7.0.