How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M NaOH solution?

Jan 4, 2018

Well, you need to know that phosphoric acid is a DIACID in aqueous solution... The third proton is rarely removed.

Explanation:

And we write the stoichiometric equation as...

${H}_{3} P {O}_{4} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} H P {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

And so .................

${n}_{N a O H} = 15 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 1.284 \cdot m o l \cdot {L}^{-} 1 = 0.01926 \cdot m o l$...and so we need HALF of this molar quantity, i.e. $9.63 \times {10}^{-} 3 \cdot m o l$, with respect to ${H}_{3} P {O}_{4}$.

And so we take the quotient....$\frac{9.63 \times {10}^{-} 3 \cdot m o l}{0.655 \cdot m o l \cdot {L}^{-} 1} = 14.7 \times {10}^{-} 3 \cdot L = 14.7 \cdot m L$ to get the volume required for equivalence.

Jan 4, 2018

$V = 9.80 m L$

Explanation:

Moles of $N a O H$ $= \frac{1.284}{1000} \times 15$

Moles of $N a O H$ $= 0.01926$

Reaction will be -
${H}_{3} P {O}_{4} + 3 N a O H \rightarrow N {a}_{3} P {O}_{4} + 3 {H}_{2} O$

So, $\frac{0.01926}{3}$ moles are required to neutralize ${H}_{3} P {O}_{4}$

Moles required to neutralize ${H}_{3} P {O}_{4} = 0.00642$

Let the volume required to neutralize ${H}_{3} P {O}_{4}$ be $V$ mL.

Moles of ${H}_{3} P {O}_{4}$ $= \frac{0.655}{1000} \times V$

$0.00642 = \frac{0.655}{1000} \times V$

$V = 0.00642 \times \frac{1000}{0.655}$

$V = \frac{6.42}{0.655}$

$V = 9.80 m L$