Question

H3PO4 + 3NaOH  Na3PO4 + 3H2O

How many milliliters of 2.2 M sodium hydroxide would be required to neutralize a 500.0 mL, 3.0 M solution of phosphoric acid?

Answer 1

We can solve this problem using the equation below

(Ma $\times$ Va) $\div$ na = (Mb $\times$ Vb)$\div$nb

Where Ma , Mb are the molarities of acid and base respectively

Va and Vb are the volume of acid and base respectively

na and nb are the number of moles of acid and base obtained from the balanced equation (that is coefficients of H3PO4 and NaOH in the balanced equation)

Equation for neutralization is given as

H3PO4 + 3NaOH $\rightarrow$ Na3PO4 + 3H2O

let us see what all values we have

Ma = 3 M

Va = 500.0 ml

na= 1 ( as coefficient in the above equation is 1)

Mb = 2.2 M

nb = 3

Vb = ?

Rearranging the equation given above for Vb we get

Vb = (Ma $\times$ Va $\times$ nb ) $\div$ (Mb $\times$ na)

= (3 $\times$ 500 $\times$ 3 ) $\div$ (2.2 $\times$ 1) = 4500 $\div$ 2.2 =2045.45 ml

Volume of sodium hydroxide required = 2045.45 ml