What is the conjugate base of HSO4-? Express your answer ...

There are various different cases for what happens when you put acids and/or bases in solutions. Everyone learns how to calculate the pH with pH=-log10([H+]) and then courses cover the use of ICE tables for weak acids/bases. Here are almost all the cases I can think of, sorted by difficulty:

(Easiest)

Trivial:

Strong acid + strong base in stoichiometric quantities
Weak acid + weak base in stoichiometric quantities, where acid's Ka and base's Kb have the same value

If the acid and the base both dissociate completely, or don't but have equal strength, and they are in equal amounts (after multiplying their concentrations if they have multiple H+/OH-), the resulting solution will always be neutral.
Example: 0.05M Ca(OH)2 + 0.10M HCl -> 0.05M CaCl2 => pH=pKw/2=7.00

Very easy:

Strong acid in any quantity
Strong base in any quantity

The acid/base dissociates to produce an anion/cation and leaves hydronium/hydroxide, so the pH is simply -1 times the logarithm of the acid's/base's initial concentration.
Example: 0.10M NaOH => pH=-log10(0.10)=13.00

Easy, with another step:

Strong acid + strong acid in any quantities
Strong base + strong base in any quantities
Strong acid + strong base in non-stoichiometric quantities

After you add the quantities together this is simply the same thing as the last group of cases. If you have two strong acids (or two strong bases), add their initial concentrations. If you have a strong acid and a strong base, subtract the concentration of one from the concentration of the other.
Example: 0.15M HCl + 0.85M HBr => pH=-log10(1.00)=0.00

Average:

Weak acid in any quantity
Weak base in any quantity

Requires an ICE table. The weak acid/base partially dissociates, leaving some HA/BOH and some H+/OH-. You'll have to do a calculation with the Ka/Kb to predict the final pH.
Example: 0.1M CH3O2H -> (x)M H+ + (0.1-x)M CH3O2-, pKa=4.76 => some handwaving => pH=2.88

Average, with an extra step:

Strong acid + weak base in stoichiometric quantities
Weak acid + strong base in stoichiometric quantities

Same as the previous cases, but first, invert the problem: assume the strong acid/base reacts with the weak base/acid and produces water, some irrelevant anions/cations, and the conjugate acid/base of the weak reactant. Then find the conjugate's equilibrium constant: pKa=14-pKb (or for the weak acid, pKb=14-pKa).
Example: 0.1M NaOH + 0.1M HF -> 0.1M Na+ + 0.1M F- => 0.1M F- -> (x)M HF + (0.1-x)M F-, pKb=14.00-3.20=10.80 => use ICE table => pH=11.88

Slightly harder, multi-step:

Strong acid that also has another weak dissociation
Strong base that also has another weak ionization
Strong acid + weak acid
Strong base + weak base
Strong acid + weak base in non-stoichiometric quantities
Strong base + weak acid in non-stoichiometric quantities

Assume the first H+/OH- dissociates completely and record the concentration. Then do an ICE table with those ions in the initial concentration, and the second ionization as an independent acid/base.
Example: 0.001M H2SO4 -> 0.001M H+ + 0.001M HSO4- -> (0.001+x)M H+ + (0.001-x)M HSO4- + (x)M SO42- => too lazy to calculate, queried wolframalpha => pH=2.70

Very difficult:

Weak acid + weak base, in non-stoichiometric quantities and/or with non-equivalent Ka and Kb
Weak acid + weak acid, with different Ka values
Weak base + weak base, with different Kb values

I don't think it's possible to do ICE tables for multiple different reactions simultaneously, where the products of each one influence the reactants of the other. I think you can approximate the solution with a series of small-step calculations, where you assume each reaction progresses a little bit and calculate the change in pH, then repeat the process until it settles around a fixed point. Something similar to the method of successive approximations, but for two processes running parallel to each other and whose products interact with each other. Perhaps you could find an exact solution with Markov chains?
Example: 0.1M CH3CO2H + 0.1M NH3; 0.1M HCN + 0.1M HCO2H; 0.1M CH3NH2 + 0.1M C5H5N
(I checked these in WolframAlpha and it doesn't know how to find the pH.)

Even harder, multi-step:

Weak acid with multiple dissociations
Weak base with multiple proton acceptances

You can't just use an ICE table for each dissociation/ionization separately because the products of one reaction with water change the concentrations of the reactants in the other aqueous reaction. For acids, the equilibrium for H2A + 2H2O <-> HA- + H3O+ + H2O <-> A2- + 2H3O+ could be literally anywhere between the first reactants and the last products (you don't know where without knowing pKa1 and pKa2) and furthermore, it could have a lot of H2A with very little HA- and A2- (where pKa1 and pKa2 are small), or lots of HA- and not much H2A and A2- (where pKa1 is large and pKa2 is small), or lots of A2- and not much H2A and HA- (if pKa1 and pKa2 are very large), or various other cases.
Examples: 0.1M H2CO3, pKa1=6.35, pKa2=10.33; 0.1M N2H4, pKb1=4.75, pKb2=15.08

Apparently not impossible:

Multiple weak acids/bases with multiple ionizations

Example: 0.1M H2CO3 and 0.1M N2H4 in the same solution

(Hardest)

I can think of others, but they're just combinations of the above categories. Anyway, how do you solve the last two sets of cases?

Edit: my title is wrong. I actually asked about a solution with a weak acid plus a weak base, in unequal amounts. But the two problems are essentially equivalent.

That is because sulfuric acid is a strong acid and completely disassociates in water. Therefore, the sulfate ion (SO2− 4) is the conjugate base of H SO− 4. The sulfate ion (SO4 2−) is the conjugate base of HSO−4. Reason : Even if it has a negative charge, it will never accept a H+ ion to form H2SO4 again. That is because sulfuric acid is a strong acid and completely disassociates in water. That’s why Sulphate Ion would be the conjugate base formed. The hydrogen sulfite, or bisulfite, ion is the ion HSO3-. It is the conjugate base of sulfurous acid, H2SO3. What is the conjugate base of hso4? HSO4- is a base since it has the ability to accept a proton but it is a conjugate base to H2SO4 since it is formed by the H2SO4 after donating a proton. Conjugate base. In a reaction of an acid in an aqueous solution, it loses a proton or Hydrogen ion. As the acid releases a proton or Hydrogen ion, it produces a compound that is one proton away or ... The conjugate base of HSO4^... chemistry. The conjugate base of H S O 4 ... Which of the following is known as hydronium ion? View solution. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid. ... What is the conjugate base of HSO4−? Express your answer as a chemical formula. What is the conjugate acid of HPO42− ? Express your answer as a chemical formula. Videos. Step-by-step answer 100% (2 rating) 01:43 2 0. Expert Answer 98% (66 ratings) Previous question Next question ... HSO4- ----> H+ + SO42-The conjugate acid of HPO42- is H2PO4-. HPO42- + H+ ----> H2PO4-PO43- is a weak conjugate base (since it is not the conjugate base of a strong acid, which has no strenth) has a charge of minus three, which means it can hold 3 H+'s.

What is the conjugate base of HSO4-? Express your answer ...

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Quick acid-base question

In 10th grade I asked a question about finding the pH of a solution with multiple weak acids, and the teacher said the explanation was too complicated for me. Five years and half a chemistry degree later, I still have no idea how to do it.

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