# Calculate the molarity of a sodium hydroxide solution if 25.00 mL 0.100 M maleic acid requires 22.10 mL of NaOH to reach the endpoint?

Jul 2, 2017

You have noted that maleic acid is a diacid.......

#### Explanation:

We always need a stoichiometric equation to assess the stoichiometry:

$\text{HO(O=)CCH=CHC(=O)OH(aq)" + "2NaOH(aq)}$

$\rightarrow \text{Na"^+""^(-)"O(O=)CCH=CHC(=O)""O"^(-)""^+"Na" +"2H"_2"O} \left(l\right)$

$\text{Maleic acid}$ is the cis isomer; whereas $\text{fumaric acid}$ is the trans isomer.

$\text{Moles of acid} = 25.00 \times {10}^{-} 3 \cdot L \times 0.100 \cdot m o l \cdot {L}^{-} 1 = 2.500 \times {10}^{-} 3 \cdot m o l .$

But because it is a diacid this is equivalent to $5.00 \times {10}^{-} 3 \cdot m o l$ with respect to $N a O H$........

And so...$\left[N a O H\right] = \frac{5.00 \times {10}^{-} 3 \cdot m o l}{22.10 \times {10}^{-} 3 \cdot L} \cong 0.2 \cdot m o l \cdot {L}^{-} 1$