The 4 questions are about titrations
Here question 1 and 2
Mixture of acids and bases reacting and what is in excess finally H+ or OH-.png
Molarity of vinegar and mass percentage .png
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
Molarity of vinegar and mass percentage .png
Vinegar is acetic acid and they reactin 1:1 molar ratio.
HC
2H
3O
2 (aq) + NaOH ---> NaC
2H
3O
2+ H2O
Planning
M*V (L) moles of NaOH --> Moles of HC
2H
3O
2-->divide by volume of vinegar (0.010L) Molarity of HC
2H
3O
2Density = mass /Volume
get the mass of 10 mL of vinegar solution
convert the moles HC
2H
3O
2 into grams
Mass% = mass of HC
2H
3O
2 *100%/mass of vinegar solution