can someone please help me??? 1.Calculate the molarity of an acetic acid solutio

Question

can someone please help me???
1.Calculate the molarity of an acetic acid solution if 76.34 mL ofthe solution is needed to neutralize 112 mL of 1.85 Msodium hydroxide. The equation for the reaction is

can someone please help me???
1.Calculate the molarity of an acetic acid solution if 76.34 mL ofthe solution is needed to neutralize 112 mL of 1.85 Msodium hydroxide. The equation for the reaction is

Explanation / Answer

CH3COOH + NaOH -> CH3COONa +H2O n(NaOH) = cv = 1.85 x 0.112 = 0.2072mol n(CH3COOH) = 0.2072mol c(CH3COOH) = n/v = 0.2072/0.07634 = 2.71M Hope this helps!

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