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.
1. To find the [OH-] in mol/L, we need to use the relationship between pH and pOH. pH + pOH = 14. Since the pH of the solution is 7.82, we can calculate the pOH as follows: pOH = 14 - pH = 14 - 7.82 = 6.18.
The [OH-] in mol/L is equal to 10^(-pOH). Plugging in the value of pOH, we have [OH-] = 10^(-6.18) = 6.18 x 10^(-7) mol/L.
Therefore, the correct answer is option c) 6.6 x 10^(-7) mol/L.
2. To determine which substance has a change in number of moles of 0.192 as the system comes to equilibrium, we need to look at the coefficients of the balanced equation.
From the balanced equation, we can see that the ratio of NH3 to O2 is 4:5. Therefore, for every 5 moles of O2 consumed, 4 moles of NH3 will be consumed.
Given that there are 0.93 moles of O2 initially and 0.77 moles of O2 at equilibrium, we can calculate the change in moles of O2: 0.93 - 0.77 = 0.16 moles.
Since the ratio of O2 to NO is 5:4, the change in moles of NO will be 4/5 times the change in moles of O2: (4/5) * 0.16 = 0.128 moles.
Therefore, the substance with a change in number of moles of 0.192 as the system comes to equilibrium is option d) NO.
3. The pH of a neutral solution is 7. At normal body temperature, Kw (the ionization constant of water) has a value of 2.42 x 10^(-14).
Since the pH of a neutral solution is 7, we know that the [H+] in mol/L is equal to 10^(-pH). Plugging in the value of pH, we have [H+] = 10^(-7).
Since Kw = [H+][OH-], and we know that in a neutral solution [H+] = [OH-], we can solve for [OH-]: [OH-] = Kw / [H+].
Plugging in the values, we have [OH-] = (2.42 x 10^(-14)) / (10^(-7)) = 2.42 x 10^(-7) mol/L.
Therefore, the pH of a neutral solution at normal body temperature is option d) 7.0.