What is 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

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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