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IceJOKER [234]
3 years ago
15

A 0.8115 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL

of solution. It required 22.07 mL of NaOH to reach the endpoint in the titration. What is the molarity of the NaOH solution?
Chemistry
1 answer:
inysia [295]3 years ago
5 0

Answer:

Molarity of NaOH solution is 1.009 M

Explanation:

Molar mass of HCl is 36.46 g/mol

Number moles = (mass)/(molar mass)

So, 0.8115 g of HCl = \frac{0.8115}{36.46}moles HCl = 0.02226 moles HCl

1 mol of NaOH neutralizes 1 mol of HCl.

So, if molarity of NaOH solution is S(M) then moles of NaOH required to reach endpoint is \frac{S\times 22.07}{1000}moles

So, \frac{S\times 22.07}{1000}=0.02226

or, S = 1.009

So, molarity of NaOH solution is 1.009 M

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<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

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\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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