Respuesta :
Answer:
0.043 M
Explanation:
The reaction that takes place is:
- Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
First we calculate how many HCl moles reacted, using the given concentration and volume required to reach the equivalence point:
- 0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺
As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.
With that in mind we can calculate the hydroxide ion concentration in the original sample solution, using the calculated number of moles and given volume:
- 1.0817 mmol OH⁻ / 25.0 mL = 0.043 M
The titration is termed the neutralization reaction with the acid and base. The concentration of hydroxide in the titration is 0.0865 M.
What is a neutralization reaction?
The neutralization reaction is given as the reaction in which the acid and base react to form the salt and water, stabilizing the pH of the solution.
The neutralization of acid and base to identify the strength can be given as:
[tex]\rm M_V_1=M_2V_2[/tex]
Substituting the strength and the volume of calcium hydroxide and the HCl with the volume:
[tex]\rm 0.029\;M\;\times\;37.3\;mL=M_2\;\times\;25\;mL\\M_2=\dfrac{ 0.029\;M\;\times\;37.3\;mL}{25\;mL} \\M_2=0.0432\;M[/tex]
The strength of the calcium hydroxide in the reaction is 0.04326 M.
One molar unit of calcium hydroxide results in 2 molar units of hydroxide. The molar unit of hydroxide in 0.04326 M calcium hydroxide is:
[tex]\rm 1\;M\;Ca(OH)_2=2\;M\;OH^-\\\\0.04326\;M\;Ca(OH)_2=0.04326\;\times\;2\;M\;OH^-\\0.04326\;M\;Ca(OH)_2=0.0865\;M\;OH^-[/tex]
The concentration of hydroxide ion in the titration is 0.0865 M.
Learn more about molarity, here:
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