Hey guys! Stuck on some pH problems! Thanks so much in advanced:
A) Calculate the pH of a 0.200 M solution of potassium acetate, KCH3CO2, in water at 25 °C. At this temperature, Ka = 1.76 × 10-5 for acetic acid.
1. 2.73 2. 5.45 3. 9.02 4. 9.94 5. 10.16 6. none of these
Maybe Answer:
Kbacetate=10^-14/1.76*10^-5=5.68*10-16
[OH-]=sqroot{5.68*10^-10*.2M)=1.060*10^-5, so pOH=4.97
ph=14-4.97=9.02, maybe answer 3
B) The pH of which salt solution will be greater than 7.00 at 25 °C? (Assume that each solution has a molar concentration of 0.20 M.)
1.NH4Cl 2. NaI 3. KNO3 4. LiHSO4 5.KHCO3
Maybe Answer:
5)KHCO3 because it is basic due to its components
C)At 25 °C, the pH of a solution of NH4Cl is 5.18. The Ka value of NH4+ is
5.65 × 10-10 at 25 °C. What is the molar concentration of the NH4Cl solution?
1. 0.012 M 4. 0.077 M
2. 0.041 M 5. 0.28 M
3. 0.059 M
Maybe Answer: 10^-5.18=[H+]=6.607*10^-6 (square this to get the top value for the Ka function because H+ and OH- are the same in this case?):
5.65*10^-10=4.365*10^-11/x2
x=.00772M=4.0.077M
D) For the diprotic acid telluric acid, H2TeO4, at 18 °C the value of is
2.09 × 10-8 and is 6.46 × 10-12. What is the molar concentration of H+ in 0.150 M H2TeO4 at 18 °C?
1.7.48 × 10-3 4. 3.12 × 10-9
2. 4.99 × 10-2 5. 6.46 × 10-12
3. 5.60 × 10-5 6. none of the previous answers
Maybe Answer:
Step 1) 2.09*10^-8=x^2/.15-x
3.135*10^-9-2.09*10^-8x-x^2=0
x=5.598*10^-5M
Step 2) 6.46*10^-12=5.598*10^-5x+x2/5.598*10^-5-x
3.616*10^-16-5.598*10^-5-x2
x=6.46*10^-12, so my answer would be 5
E) Calculate the percent dissociation for a 1.00 × 10-6 M solution of HCN at 25 °C.
Ka = 4.90 × 10-10.
Maybe answer:
HA->H++A-
1.00*10^-6M-x x x
4.90*10^-10=x2/1.00*10^-6-x
4.9*10^-16-4.90*10^-10x-x2=0
x*100=percent dissociation (?)
2.19*10^-6%
Any help is appreciated! I worked really hard on this! Thanks so much!!!!
A. 4.97 and 9.02 are right but I obtained 1.066E-5 and not 1.060E-5.
B. ok for KHCO3 but I think your reasoning why is weak. The moon is made of green cheese, too, because of its components. The problem doesn't ask for a reason but the answer is because of the hydrolysis of the HCO3^- (and of course that's because of the components).
C. ok
D. I don't know where you're going on D. First, the second ionization never enters into the process since it is so much smaller than k1. Use k1 and stop there; i.e., H^+ = 5.598E-5 is the right answer (rounded of course to the correct number of s.f.).
Thanks for the help! Sorry about D; I've never experiences a problem like that before. I guess I have a question about D. I understand that k2 is a lot smaller than k1, but since you get a different answer when doing the second ionization doesn't that mean that it is still significant enough to impact the H+ concentration? I'm just confused on that. And if you get a chance, could you look over E? That would be a lot of help! Thanks again! I appreciate it!
On D. With k1 = about 10^-8 and k2 = about 10^-12, that is about 10,000 less for K2 so whatever (H^+) is from k1, the contribution from k2 is 10,000 less than that. So with 5.6E-5 for k1, trying to add a 10,000th of that is fruitless. I think part of your confusion stems from your statement that the H^+ from K2 is significant. It isn't and your calculation doesn't show that. Here is how you calculate H^+ from k2. In fact your answer for the k2 part is 6.46E-12 which is correct. I think you just added it in wrong. And you can shorten the way you calculate the contribution from k2.
HTeO4^- --> H^+ + TeO4^2-
The contribution of (H^+) by k2 = (TeO4^2-) so you need to find (TeO4^2-).
k2 = 6.46E-12 = (H^+)(TeO4^2-)/HTeO4^-
But your calculation from k1 (where you determined the (H^+) from k1) tells you that (H^+) = (HTeO4^-). So (H^+) in the numerator of k2 cancels with (HTeO4^-) in the denominator of k2 which makes (TeO4^2-) = k2 = 6.46E-12. So total (H^+) from the system is 5.6E-5 + 6.5E-12 = 5.6E-5 which shows very clearly that the H^+ from k2 really is quite negligible. Best of luck.
A) To calculate the pH of a solution of potassium acetate, KCH3CO2, you need to consider the hydrolysis reaction of the acetate ion, CH3CO2-. The hydrolysis reaction of CH3CO2- with water can be written as follows:
CH3CO2- + H2O ↔ CH3COOH + OH-
The equilibrium constant, Ka, is given as 1.76 × 10^-5 for acetic acid, CH3COOH. Since acetic acid is a weak acid, it partially dissociates and produces H+ ions. However, in this case, we are looking at the hydrolysis of CH3CO2-, which means the reverse reaction is favored and OH- ions are produced.
First, calculate the concentration of hydroxide ions, [OH-], by using the concentration of KCH3CO2, which is 0.200 M.
[OH-] = √(Ka * [KCH3CO2])
[OH-] = √(1.76 × 10^-5 * 0.200) = 1.060 × 10^-5 M
Next, calculate the pOH using the concentration of hydroxide ions:
pOH = -log([OH-]) = -log(1.060 × 10^-5) = 4.97
Finally, calculate the pH using the relationship:
pH = 14 - pOH = 14 - 4.97 = 9.03
So, the pH of the 0.200 M solution of potassium acetate is approximately 9.03. The correct answer is 3. (9.02).
B) To determine the salt solution with a pH greater than 7.00 at 25 °C, you need to consider the ions produced when the salt dissolves in water.
1) NH4Cl: When NH4Cl dissolves, it produces NH4+ and Cl- ions. NH4+ can act as a weak acid, causing the solution to be acidic.
2) NaI: When NaI dissolves, it produces Na+ and I- ions. Neither of these ions has any acidic or basic properties, so the solution will be neutral.
3) KNO3: When KNO3 dissolves, it produces K+ and NO3- ions. Neither of these ions has any acidic or basic properties, so the solution will be neutral.
4) LiHSO4: When LiHSO4 dissolves, it produces Li+ and HSO4- ions. HSO4- can act as a weak acid, causing the solution to be acidic.
5) KHCO3: When KHCO3 dissolves, it produces K+ and HCO3- ions. HCO3- can act as a weak base, causing the solution to be slightly basic.
Based on the explanation above, the correct answer is 5. (KHCO3).
C) To determine the molar concentration of the NH4Cl solution given the pH, you need to consider the dissociation of NH4+ in water.
NH4+ acts as a weak acid and can dissociate as follows:
NH4+ + H2O ↔ NH3 + H3O+
The equilibrium constant, Ka, is given as 5.65 × 10^-10 for NH4+.
Using the given pH of 5.18, calculate the concentration of H+ ions:
[H+] = 10^(-pH) = 10^(-5.18) = 6.607 × 10^-6 M
Since NH4+ dissociates into 1 NH3 molecule and 1 H+ ion, the molar concentration of NH4+ can be assumed to be equal to the concentration of H+ ions. Therefore, the molar concentration of the NH4Cl solution is 6.607 × 10^-6 M.
D) To determine the molar concentration of H+ in a solution of H2TeO4, you need to consider the dissociation constants, Ka1 and Ka2, for the two ionizable protons.
The dissociation of H2TeO4 can be written as follows:
H2TeO4 ↔ H+ + HTeO4-
HTeO4- ↔ H+ + TeO4^2-
The equilibrium constants, Ka1 and Ka2, are given as 2.09 × 10^-8 and 6.46 × 10^-12, respectively.
First, calculate the concentration of H+ ions from the dissociation of H2TeO4 using the initial molar concentration of 0.150 M:
[H+] = √((Ka1 * Ka2 * [H2TeO4]) / ([H2TeO4] + Ka1 * [H2TeO4] + Ka1 * Ka2))
[H+] = √((2.09 × 10^-8 * 6.46 × 10^-12 * 0.150) / (0.150 + 2.09 × 10^-8 * 0.150 + 2.09 × 10^-8 * 6.46 × 10^-12))
[H+] = 5.598 × 10^-5 M
So, the molar concentration of H+ ions in the 0.150 M H2TeO4 solution is 5.598 × 10^-5 M. The correct answer is 5. (6.46 × 10^-12).
E) To calculate the percent dissociation for a solution of HCN, you need to consider the dissociation of the weak acid, HCN, in water.
HCN ↔ H+ + CN-
The equilibrium constant, Ka, is given as 4.90 × 10^-10 for HCN.
Given the initial molar concentration of 1.00 × 10^-6 M for HCN, let's assume that x is the molar concentration of H+ and CN- ions formed due to dissociation.
Using the equilibrium expression for the dissociation, we have:
Ka = [H+][CN-] / [HCN] = [x][x] / [1.00 × 10^-6 - x]
Setting up the equation and solving for x:
4.90 × 10^-10 = x^2 / (1.00 × 10^-6 - x)
By assuming that x is small compared to the initial concentration of HCN (1.00 × 10^-6), we can simplify the equation as:
4.9 × 10^-10 ≈ x^2 / (1.00 × 10^-6)
x ≈ √(4.9 × 10^-10 * 1.00 × 10^-6) ≈ 2.21 × 10^-8 M
The percent dissociation is then calculated by:
Percent dissociation = (x / [HCN]) * 100 = (2.21 × 10^-8 / 1.00 × 10^-6) * 100 ≈ 2.21 × 10^-3%
So, the correct answer for the percent dissociation of the 1.00 × 10^-6 M HCN solution is approximately 2.21 × 10^-3%.