The 4 questions are about titrations
Here question 1 and 2
 

For first question
You calculate H+ ions moles from given information
Moles of HCl = M* V (L) = moles of H⁺
Moles of HNO₃ = M* V (L) = moles of H⁺
Add them  moles of H⁺ = 0.0300 moles
Similarly calculate moles from all bases.
Moles of RbOH = M*V(L)= moles OH ^-
Moles Ca(OH)2 = M*V (L) and double of this is equal to OH ^- moles
Add moles of OH^- from both bases = .0447 moles
Now H⁺ + OH^- --> H₂O
1:1 molar ratio
Now we can see OH- is more than H+
So solution is basic
Excess of OH- = .0447-.0300= .0147 moles
[OH-] = total moles left / total volume in L

Quote from: Siba on December 07, 2025, 05:40:37 PMThe 4 questions are about titrations
Here question 1 and 2
 

Vinegar is acetic acid and they reactin 1:1 molar ratio.
HC₂H₃O₂ (aq) + NaOH ---> NaC₂H₃O₂+ H2O
Planning
M*V (L) moles of NaOH --> Moles of HC₂H₃O₂-->divide by volume of vinegar (0.010L) Molarity of HC₂H₃O₂

Density = mass /Volume
get the mass of 10 mL of vinegar solution
convert the moles HC₂H₃O₂ into grams
Mass% = mass of HC₂H₃O₂ *100%/mass of vinegar solution