Question

You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. calculate the moles of Hcl and H2so4 required to neutralize the Naoh solution

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Answer 1

`n_(HCl)=0.2molHCl\\n_(H_2SO_4)=0.1molH_2SO_4`

Step-by-step explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

`NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O`

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

`n_(HCl)=n_(NaOH)=V_(NaOH)M_(NaOH)=0.01L*20mol/L=0.2molHCl\\\\2n_(H_2SO_4)=n_(NaOH)\\\\n_(H_2SO_4)=(1)/(2) V_(NaOH)M_(NaOH)=(0.01L*20mol/L)/(2) =0.1molH_2SO_4`

Best regards!