standardization Data Trial 1 Trial 2 Trial 3 mass KHP (g) 0.5164 0.5182 05143 Vo

Question

standardization Data Trial 1 Trial 2 Trial 3 mass KHP (g) 0.5164 0.5182 05143 Volume base (ml) 1189 11.81 B 1186 B 0.2127 M 0.2149 M 0.2123 M 0.2133 M Average these values Vinegar Data for acetic acid vinegar (m) 500 5.00 5.00 volume base (mu) 21.49 B 20.10 a 20.80 0.887 M a Acid Molarity 0.917 0.857 M 0.887 M 5.51 5.33% 5.33 5.15 Unknown Data for molarity Hx Unknown Number 48 i vol unknown 10.00 E 1000 10.00 E 1p4d2noc 3x954 Volume base (mL) 9.18 10.72 1008 3 Molarity 0.1958 M 0.2286 M 0.2150 M 0.2131 M Average these values Problem: calculate the volume (mL) of 0.3667 M nitric acid required to just d i neutralize 6.23 g calcium hydroxide

Explanation / Answer

Ca(OH)2 + 2HNO3---------------> Ca(NO3)2 + 2H2O

1 mole        2 moles

no of moles of Ca(OH)2 = W/G.M.Wt   = 6.23/74 = 0.084 moles

from balanced equation

1 moles of Ca(OH)2 react with 2 moles of HNO3

0.084 moles of Ca(OH)2 react with = 2*0.084/1 = 0.168 moles of HNO3

no of moles of HNO3 = molarity * volume in L

   0.168                       = 0.3667* volume in L

volume in L                  = 0.168/0.3667   = 0.458L = 458ml

volume of HNO3 = 458ml

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