A 20.00 mL Ba ( OH ) 2 solution of unknown concentration was neutralized by the addition of 42.50 mL of a 0.1255 M HCl solution. Write the balanced molecular equation for the neutralization reaction between HCl and Ba ( OH ) 2 in aqueous solution. Include physical states. molecular equation: Ba^{2+}(aq) +2OH^{-}(aq) +H^{+}(aq) +Cl^{-}(aq)<=>H_{2}O(l) +Ba^{2+}(aq) +Cl^{-}(aq) Ba 2 + ( aq ) + 2 OH − ( aq ) + H + ( aq ) + Cl − ( aq ) − ⇀ ↽ − H 2 O ( l ) + Ba 2 + ( aq ) + Cl − ( aq ) Calculate the concentration of Ba ( OH ) 2 in the original 20.00 mL solution. [ Ba ( OH ) 2 ] = M Calculate the concentrations of Ba 2 + and Cl − in solution following the neutralization reaction.

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Аноним Аноним · Answer 1 · 2020-02-28T09:31:16+01:00

Answer: The concentration of [tex]Ba^{2+}\text{ and }Cl^-[/tex] ions in the solution are 0.0426 M and 0.0852 M respectively

Explanation:

Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce a salt and water molecule.

The chemical equation for the reaction of HCl and barium hydroxide follows:

[tex]2HCl(aq.)+Ba(OH)_2(aq.)\rightarrow BaCl_2(s)+2H_2O(l)[/tex]

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=?M\\V_1=20.00mL\\n_2=2\\M_2=0.1255M\\V_2=42.50mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 20.00=1\times 0.1255\times 42.50\\\\M_1=\frac{1\times 0.1255\times 42.50}{1\times 20.00}=0.267M[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)

Molarity of HCl solution = 0.1255 M

Volume of solution = 42.50 mL = 0.04250 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.1255M=\frac{\text{Moles of HCl}}{0.04250L}\\\\\text{Moles of HCl}=(0.1255mol/L\times 0.04250L)=5.33\times 10^{-3}mol[/tex]

By Stoichiometry of the reaction:

2 moles of HCl produces 1 mole of barium chloride

So, [tex]5.33\times 10^{-3}mol[/tex] of HCl will produce = [tex]\frac{1}{2}\times 5.33\times 10^{-3}mol=2.665\times 10^{-3}mol[/tex]

Moles of barium chloride = [tex]2.665\times 10^{-3}mol[/tex]

Volume of solution = [20.00 + 42.50] mL = 62.5 mL = 0.0625 L

Putting values in equation 1, we get:

[tex]\text{Molarity of barium chloride}=\frac{2.665\times 10^{-3}}{0.0625}=0.0426M[/tex]

1 mole of barium chloride produces 1 mole of barium ions and 2 moles of chloride ions

So, concentration of barium ions = 0.0426 M

Concentration of chloride ions = (2 × 0.0426) = 0.0852 M

Hence, the concentration of [tex]Ba^{2+}\text{ and }Cl^-[/tex] ions in the solution are 0.0426 M and 0.0852 M respectively