Question

What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH

Answer 1

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

$HCl + NaOH \rightarrow NaCl + H_2O$

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

$M_1\times V_1(NaOH)= M_2\times V_2(HCl)$

$0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL$

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

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