I agree with 2 and 3 but not 1.
I didn't calculate all the way for the acids but the answer you have for a I don't think is right. Neither is the answer for c. b and d are OK.
1. Which solution has the lowest pH?
a. 0.0045 M C6H5COOH (Ka=6.3x10-5)
b. 0.00018 M HNO3
c. 0.016 M H2CO3 (Ka1=4.3x10-7) ***
d. 0.00075 M HCOOH (Ka=1.8x10-4)
*The pH I calculated: a. 2.3, b. 3.7, c. 1.8, d. 3.1
2. In the equilibrium, F-(aq)+HO2CCO2H(aq)⇄HF(aq)+HO2CCOO-(aq), the conjugate base is
a. F- ***
b. HO2CCO2H
c. HF
d. HO2CCOO-
3. Which of the following is a Brønsted-Lowry acid-base reaction?
a. 2Na(s)+2H2O(l)⇄2NaOH(aq)+H2(g)
b. NH4^+(aq)+H2O(l)⇄NH3(aq)+H3O^+(aq) ***
c. NaHCO3(aq)+2HCl(aq)⇄NaCl(aq)+H2O(l)+CO2(g)
d. AgNO3(aq)+NaCl(aq)⇄AgCl(s)+NaNO3(aq)
2 answers
oops. On #2, F^- is the base in the reaction. The CONJUCATE base is d.
In #3. the equation you have written is not correct. It isn't balanced. If it were balanced it would be
NaHCO3 + HCl ==> H2CO3 + 2NaCl in which case the HCO3]^- accepts a proton from the HCl so the HCl is an acid and the [HCO3]^- is a base in the B and L theory. Now if we take your equation and say that the 2HCl is just extra HCl, so that's a B and L acid/base also with some extra HCl thrown in. In the above reaction I wrote, the H2CO3 is not stable and gives => H2O + CO2. If you write the intermediate step I think you call it a Bronsted-Lowry acid/base rxn. If you don't go through the intermediate step I think it is not.
In #3. the equation you have written is not correct. It isn't balanced. If it were balanced it would be
NaHCO3 + HCl ==> H2CO3 + 2NaCl in which case the HCO3]^- accepts a proton from the HCl so the HCl is an acid and the [HCO3]^- is a base in the B and L theory. Now if we take your equation and say that the 2HCl is just extra HCl, so that's a B and L acid/base also with some extra HCl thrown in. In the above reaction I wrote, the H2CO3 is not stable and gives => H2O + CO2. If you write the intermediate step I think you call it a Bronsted-Lowry acid/base rxn. If you don't go through the intermediate step I think it is not.
1 answer