you have 0.1 Moles of sulfuric acid with 0.2 moles of H ions thus requiring 0.2 moles of NaOH - molecular weight = 40 so 40 x 0.2 = 8g NaOH
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
Balanced equation. H2SO4 + 2NaOH >> Na2SO4 + 2H2O 10 grams NaOH (1mol NaOH/39.998g )(1mol H2SO4/2mol NaOH )(98.086g H2SO4/1molH2SO4 ) = 12.26 grams of H2SO4
112,64 g of NaOH and 138,1 g of H2SO4
14.1 mL is required to titrate 10.00 ml of 0.526 M H2SO4.
(25.00ml H2SO4)(H2SO4 M) = (22.65ml NaOH)(0.550M) = 0.4983M H2SO4
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
Balanced equation. H2SO4 + 2NaOH >> Na2SO4 + 2H2O 10 grams NaOH (1mol NaOH/39.998g )(1mol H2SO4/2mol NaOH )(98.086g H2SO4/1molH2SO4 ) = 12.26 grams of H2SO4
112,64 g of NaOH and 138,1 g of H2SO4
14.1 mL is required to titrate 10.00 ml of 0.526 M H2SO4.
(25.00ml H2SO4)(H2SO4 M) = (22.65ml NaOH)(0.550M) = 0.4983M H2SO4
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
40 grams, this is the 1M NaOH standard laboratory solution.
Both NaOH and H2SO4 have no colour, even not after complete reaction (Na2SO4 and H2O are also colourless)
Vinegar
Two parts.Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters )0.5000 M NaOH = X moles/0.1 L= 0.05 moles NaOH--------------------------------now,0.05 moles NaOH (39.998 grams/1 mole NaOH)= 1.99 grams NaOH required====================( could call it 2.00 grams )
8 grams NaOH (1 mole NaOH/39.998 grams) = 0.2 moles NaOH
Methyl orange