Question

If 19.5mL of a 0.05M solution of Na2CO3 is neutralized with 20mL of HCl, what is the concentration of the HCl?

Answer 1

`color(blue)(M_2)=0.05M`

Explanation:

Formula for Dilution
`color(red)(M_1*V_1)=color(blue)(M_2*V_2)`

Convert mL to L, and M to mol.
However, we know that M=mol.
So, 0.05 M=0.05mol `Na_2CO_3` and this is your `color(red)M_1`

For `color(red)(V_1)`,
`19.5cancel(mL)*(1L)/(1000cancel(mL))=0.0195 L`

And for `color(blue)(V_2)`,
`20.0cancel(mL)*(1L)/(1000cancel(mL))=0.02L`

`color(blue)(M_2)=color(red)((M_1*V_1))/color(blue)(V_2)`
`color(blue)(M_2)=color(red)(0.05molNa_2CO_3*0.0195cancel(L))/(color(blue)(0.02cancel(L))`
`color(blue)(M_2)=0.05M`