Acid-Base Titration Report Sheet A. Acetic Acid in Vinegar A.2 Molarity (M) of N

Question

Acid-Base Titration Report Sheet A. Acetic Acid in Vinegar A.2 Molarity (M) of NaOH (stated on label) Volume ,50m on label) Trial 1 Trial 2 Trial 3 A.3 A.4 A.5 Initial NaOH level in buret Final NaOH level in buret Volume (mL) of NaOH used Average volume (mL) 11300| 21.1mL IL A.6 Average volume in liters ( A.7 Moles of NaOH used in titration mole NaOH A8 Moles of HC,H,02 neutralized by NaOH mole HC H,O A.9 Molarity of HC H,O M HC H,0 A.10 Grams HC H O HC H,O Show calculations A.11 Percent (m/v) HC,H,O, in vinegar -# HC,H,O Show caleulations Questions and Probiems Q1 How many grams of Mg(OH), will be needed to neutralize 25 mlL of stomach acid if stomach acid is 0.10 M HCI Q2 How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H PO solution?

Explanation / Answer

Mg(OH)2(s)+2HCl(aq)MgCl2(aq)+2H2O(l)

1mole is enough to neutralize 2 moles of HCl

as 0.10 moles of HCl is present 25 mi of solution , no of moles present in 1 lt of solution 1 L=103 mL of solution,

(25mL* 0.10 moles HCl ) / 103mL = 0.0025 moles HCl

so to nuetralize completely  0.0025moles HCl

(0.0025moles HCl * 1 mole Mg(OH)2) / 2moles HCl = 0.00125 moles Mg(OH)2

3 NaOH(aq) + H3PO4(aq) Na3¬PO4(aq) + 3 H2O(l)

3 mole is enough to neutralize 1 moles of H3PO4

for exactly no of moles H3PO4 = 0.20 M x 0.015 L = 0.0030 mol H3PO4

no of moles of NaOH = 0.0030 mol H3PO4 x (3 mol NaOH/1 mol H3PO4) = 0.0090 mol NaOH

Molarity = no of moles of NaOH/L

L = no of moles of NaOH /Molarity = 0.0090/0.10 = 0.090 L or 90. mL

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