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Kay [80]
3 years ago
12

The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applied to an acid-base tritration. The equivalence point occurs when 29.45 mL o

f 0.2343 M NaOH is added to a 38.54 mL portion of H2SO4. Calculate the original [H2SO4].
Chemistry
1 answer:
Svet_ta [14]3 years ago
8 0
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applies to an acid-base tritration.
            moles NaOH = c · V = 0.2343 mmol/mL · 29.45 mL = 6.900135 mmol
            moles H2SO4 = 6.900135 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.4500675 mmol
Hence
            [H2SO4]= n/V = 3.4500675 mmol / 38.54 mL = 0.08952 M
The answer to this question is  [H2SO4] = 0.08952 M

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Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

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\Delta T = change in temperature = 0.360 K

Now put all the given values in the above formula, we get:

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

Now put all the given values in the above formula, we get:

\Delta H=\frac{460.8J}{0.0268mole}

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