A 20.0 mL sample of vinegar (dilute acetic acid. CH_3CO_2H) is titrated and foun

Question

A 20.0 mL sample of vinegar (dilute acetic acid. CH_3CO_2H) is titrated and found to react with 94.7 ml, of 0.200 M NaOH. What is the molarity (M) of the acetic acid solution? The reaction is NaOH(aq) + CH_3CO_2H(aq) rightarrow CH_3CO_2Na(aq) + H_2O(l) When 50.0 mL of 0.200 MAgNO_3 and 50.0 mL of 0.200 M HCI are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 23.2 degree C to 24.8 degree C. The temperature increase is caused by the following reaction: AgNO_3(aq) + HC1 rightarrow AgCl(s) + HNO_3(aq) Calculate the enthalpy change (delta H) for this reaction in kJ/mol AgNO_3. (Assume that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g middot degree C)

Explanation / Answer

(3)

Neutraisation formula,

M1V1/n1 = M2V2/n2

M1*20.0 / 1 = 0.200 * 94.7 / 1

M1 = Molarity of Acetic acid = 0.947 M

(4)

Heat change, q = m * s (t2 - t1)

q = 100 * 4.18 * (24.8 - 23.2)

q = 668.8 J

Moles of AgNO3 = Molarity * volume of solution in L = 0.200 * 50.0 / 1000 = 0.0100 mol

Therefore,

enthalpy change, deltaH = q / n = 668.8 / 0.0100 = 66880 J = 669. kJ

So, 669. kJ of heat is liberated.

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