Respuesta :
Answer: The [tex][OH^-][/tex] of a solution is [tex]10^{-12}[/tex] M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]HCl[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.0025\times 1000}{250}=0.01[/tex]
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl\rightarrow H^++Cl^{-}[/tex]
According to stoichiometry,
1 mole of [tex]HCl[/tex] gives 1 mole of [tex]H^+[/tex]
Thus [tex]0.01[/tex] moles of [tex]HCl[/tex] gives =[tex]\frac{1}{1}\times 0.01=0.01[/tex] moles of [tex]H^+[/tex]
Putting in the values:
[tex][H^+][OH^-]=10^{-14}[/tex]
[tex][0.01][OH^-]=10^{-14}[/tex]
[tex][OH^-]=10^{-12}[/tex]
Thus the [tex][OH^-][/tex] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M
The [OH-] of a solution is [tex]10^{12}[/tex] M.
What is Molarity?
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
M = n/ V..................(1)
where,
n = moles of solute
V = volume of solution in ml
Calculation for number of moles:
Moles of HCl = 0.0912 g/ 36.5 g/mol = 0.0025 mol
On substituting the values in equation 1:
M = n/ V
M= 0.0025*1000 / 250
M=0.01 M
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl---- > H^++Cl^-[/tex]
According to stoichiometry,
1 mole of HCl gives 1 mole of [tex]H^+[/tex]
Thus, 0.01 moles of HCl gives = 1 / 1 *0.01 = 0.01 mole of [tex]H^+[/tex]
On adding the values:
[tex][H^+][OH^-]=10^{14}\\\\(0.01)[OH^-]=10^{-14}\\\\OH^-=10^{-12}[/tex]
Thus, the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M.
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