Write and balance the equation.
H3AsO4 + KOH ==> KH2AsO4 + H2O
mols H3AsO4 = M x L = ?
Now use the coefficients in the balanced equation to convert mols H3AsO4 to mols KOH. That's 1:1 so mols H3AsO4 = mols KOH.
Then M KOH = mols KOH/L KOH. You have mols KOH and M KOH, solve for L KOH and convert to mL if that's what you need. Note that this is the volume required to neutralize the first H of H3AsO4. It will take that same amount to neutralize the second H and that same amount again to neutralize the third H.
Can someone please answer one of these question so I would be able to know how to do the others?
1. What volume of 0.275 M KOH (aq) must be added to 75.0 mL of 0.137 M H3AsO4 (aq) to reach (a) the first stoichiometric pony; (b) the second stoichiometric point
2. What volume of 0.0848 M HCL must be added to 88.8 mL of 0.233 M Na3PO4 to reach (a) the first stoichiometric point; (b) the second stoichiometric point; (c) the third stoichiometric point
3. What volume of 0.255 M HNO3 must be added to 35.5 mL of 0158 M Na2HPO3 to reach (a) the first stoichiometric point; (b) the second stoichiometric point
4. What volume of 0.123 M NaOH(aq) must be added to 125 mL of 0.197 M H2SO3(aq) to reach (a) the first stoichiometric point; (b) the second stoichiometric point
Thank you so much!
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