# What volume of 0.170{\rm M} {\rm NaOH} is needed to neutralize 6.00{\rm g} of benzoic

I need to find the volume needed to neutralize 6.00 g of Benzoic acid with .170 M of NaOH. I figured that 122.12g = 1 M of Benzoic acid, therefore 6g/122.12= .04913 M of Benzoic acid, but i'm not sure how to figure that out compared to the volume of NaOH

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 Jul 22, 2012, 10:42 PM
Quote:
 6g/122.12= .04913 M
This should be moles, not molar.

You have 0.04913 mol of C6H6COOH

Do you know the equation for the reaction between benzoic acid and NaOH? If so, then you can proceed forward.

You can either use the formula, or use proportions.

Formula: $\text{Moles} = \text{Concentration}\ \times\ \text{Volume}$

Proportions:
0.170 mol of NaOH is in 1 L.

1 mol of NaOH is in 1/0.170 L.

x mol of NaOH is in $x$$\frac{1}{0.170}$$$ L.

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