Chemistry Q's

[grami001]

Hi everybody.... long time reader, first time posting. I had a couple of questions in Chemistry.

1. At 25C, only 1.9 g CaSO4 will dissolve in 2.00L of water. What is the equilibrium constant for the reaction below?
CaSO4 (s) <-----> Ca 2+ (aq) + SO4 2- (aq)
a. 4.9 x 10 -5
b. 1.9 x 10 -4
c. 1.4 x 10 -2
d. 7.0 x 10 -3
e. 0.90

2. A mixture of 0.200 mol NO2 and 0.200 mol CO is placed in a 1.00 L flask and given time to equilibrate. Analysis of the equilibrium mixture indicates that 0.134 mol of CO2 is present. Calculate Kc for the rxn.
NO2 (g) + CO(g) <-----> NO(g) + CO2(g)
a. 0.27
b. 0.45
c. 0.67
d. 2.0
e. 4.1

3.Nitrosyl bromide decomposes according to the chemical equation below.
2 NOBr (g) <----> 2NO (g) + Br2 (g)
When 0.260 atm of NOBr is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr decomposes. What is the equilibrium constant, Kp, for the rxn?
a. 2.3x10-3
b. 4.5x10-3
c. 3.5x10-2
d. 4.8x10-2
e. 8.0x10-2

4. Which is not an amphiprotic species in water?
a. HC2O4 -
b. CH3CO2 -
c. HPO4 2-
d. H2PO4 -
e. HCO3 -

5. An aqueous solution with a pH of 10.60 is diluted from 1.0L to 1.5L. what is the pH of the diluted solution?
a. 4.84
b. 7.07
c. 9.60
d. 10.42
e. 10.78

6. Which salt forms a 0.10M aqueous solution with the lowest pH?
a. MgCO3
b. NaCl
c. NaF
d. NH4Cl
e. K3PO4

7. What is the H30+ concentration in 0.45 M HCN (aq)? (Ka of HCN = 4.0x10-10)
Answer: 1.3x10-5 (tired of typing all the answers lol)

8. The pH of aqueous 0.30 M hypochlorous acid, HClO, is 3.99. What is the pKa of this acid?
Answer: 6.54
Thanks

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[grami001]

Anyone know at least one of them? 🙄

 

[ddswillbeback]

sorry I don't have time now to do other questions, but here is the answers for #4 and 6:

#4)b, because it doesn't have any acidic proton.
#6)d is the answer. a, c, e are basic solution since they contain an anion of a weak acid (their cations are form strong base which are neutral). B is a neutral salt since it contains a cation of strong base and an anion of a strong acid.

 

[prydA]

1) A. 1.9 g CaSO4/ 136 g/mol = mol CaSO4/2 L = M CaSO4 =x. Equilibrium constant = [x][x], so SQRT(x) = 4.879x10-5

2) E. The expression is [NO][CO2]/[NO2][CO]. Plug in (0.134)(0.134)/(0.066)(0.066) and you get 4.122.

3) A. set up an equilibrium table. the Kp expression will be Kp = [NO]^2[Br2]/[NOBr]^2. At equilibrium, there will be 0.260atm x 78%=.2086atm of NOBr. Set up the equation 0.260-2x=0.2086; x= 0.0286; [2x]^2[x]/[0.2028]^2=2.275x10-3

4) B. it cannot either gain a proton or lose a proton, which an amphoteric species is by definition supposed to be able to do

5) E. 10.6 = -log[H], [H] = 2.5x10-11; 2.5x10-11/1.5L = 1.67x10-11, -log[1.67x10-11] = 10.776

6) D. NH4Cl --> NH4OH + HCl; HCl is a strong acid, while NH4OH is a weak base

7) 4.0x10-10 = [H+][CN-]/[HCN] --> [x][x]/[0.45]; x=1.34x10-5

8) i dunno how to do this one 😡


damn that took foreverr

 

[grami001]

thanks ddswillbeback 🙂

 

[grami001]

thanks pryDa🙂

 

[osimsDDS]

hey can someone help with #8, i think i have the right idea but the numbers arent clicking!!! thanks

8. The pH of aqueous 0.30 M hypochlorous acid, HClO, is 3.99. What is the pKa of this acid?
Answer: 6.54

so basically: Ka= [H+]x[ClO-]/[HClO]

[H+] = something like 1.2x10^-4 = ph=3.99...I am estimating without calculator...
So if [H+] concentration is 1.2x10^-4 the [ClO-] is the same right??

so basically just plug is what we have to find Ka:

Ka=[1.2x10^-4]x[1.2x10^-4]/.3M
ka=4.8x10^-8

Then to find pKa you just take the -log Ka....which should be like 7.5 about but the answer is 6.54...so i dont know what I am doing wrong. thanks

 

[baedero1]

#8.

HClO + H2O ----> H3O+ + ClO-
initial = 0.3M, 0, 0
change = -x, 1.02*10^-4, +x
eq = 0.2998, 1.02*10^-4, 1.02*10^-4

Ka=(1.02*10^-4)(1.02*10^-4)/0.2998 = 3.47*10^-8

pKa=-log(3.47*10^-8)=7.459

 

[osimsDDS]

yea but the answer is 6.54, I had the same set-up as you but the answer is off by 1 I dunno whats goin on...