Question

7. If a waste container has 87.0 mL of 0.234 M acetic acid, how many milliliters of 2.00 M sodium

hydroxide is needed to neutralize the acid?​

Answer 1

`V_(base)=10.2mL`

Step-by-step explanation:

Hello.

In this case, since the neutralization of the acid requires equal number of moles of both acid and base:

`n_(acid)=n_(base)`

Whereas we can express it in terms of concentrations and volumes:

`M_(acid)V_(acid)=M_(base)V_(base)`

Thus, we can compute the volume of sodium hydroxide (base) as follows:

`V_(base)=(M_(acid)V_(acid))/(M_(base)) \\\\V_(base)=(87.0mL*0.234M)/(2.00M)\\ \\V_(base)=10.2mL`

Best regards.