Question

The volume of 0.1n hcl required to neutralise completely 2g of equimolar mixture

Answer 1

Explanation:

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Sodium Carbonates+2

Calculate the volume of 0.1M HCL required to neutralize 1g equimolar mixture of NaHCO3 and Na2CO3.?

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1 ANSWER

Nissim Raj Angdembay

Nissim Raj Angdembay, A Levels (2016)

Answered May 27, 2015 · Author has 595 answers and 2.2m answer views

An equimolar mixture is something that has equal number of moles per compound in a mixture. So, we need to find such a number of mole that when we add them together, we get 1 g mixture of Sodium Bicarbonate and Sodium Carbonate. Let that mole be x. Now, the molar mass of sodium bicarbonate times x plus the molar mass of sodium carbonate time x should give us 1 gram.

From formula, we have:

∑(n×Ar)=m∑(n×Ar)=m

x×84.0+x×106=1x×84.0+x×106=1

84x+106x=184x+106x=1

190x=1190x=1

x=1/190x=1/190

x=5.26×10−3molx=5.26×10−3mol

Since they both are present in equal number of moles, there are 5.26×10−3mol5.26×10−3mol of NaHCO3NaHCO3 and Na2CO3Na2CO3.

Now, write the balanced equations for the two:

2HCl+Na2CO3→2NaCl+H2O+CO22HCl+Na2CO3→2NaCl+H2O+CO2

HCl+NaHCO3→NaCl+H2O+CO2HCl+NaHCO3→NaCl+H2O+CO2

So,

2 mol of HClHCl reacts with 1 mol of Na2CO3Na2CO3

10.5×10−310.5×10−3 mol of HClHCl reacts with 5.26×10−35.26×10−3 mol of Na2CO3Na2CO3

1 mol of HClHCl reacts with 1 mol of NaHCO3NaHCO3

5.26×10−35.26×10−3 of HClHCl reacts with 5.26×10−35.26×10−3 mol of NaHCO3NaHCO3

So in total, you need 5.26×10−35.26×10−3 + 10.5×10−310.5×10−3 = 15.7×10−315.7×10−3 mol of HClHCl to neutralize the mixture,

Since,

n=c×vn=c×v

nc=vnc=v

v=15.7×10−30.1v=15.7×10−30.1

v=0.157dm3v=0.157dm3

or

v=0.157lv=0.157l

or

v=157ml.