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

Answer:

$V_{base}=10.2mL$

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:

$V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{87.0mL*0.234M}{2.00M}\\ \\V_{base}=10.2mL$

Best regards.