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rt APart complete Calculate [H+] for [OH−]=4.1×10−4 M. Express your answer using two significant figures. [H+][ H + ] = 2.4×10−11 M Previous Answers Correct The formula for the ion-product constant of water, Kw, is [OH−][H+]= 1×10−14. When rearranged to solve for [H+], it becomes [H+]===Kw[OH−]1.0×10−144.1×10−42.4×10−11M Part BPart complete Calculate [H+] for [OH−]=8.5×10−9 M. Express your answer using two significant figures. [H+][ H + ] = 1.2×10−6 M Previous Answers Correct Part CPart complete Calculate [H+] for a solution in which [OH−] is 100 times greater than [H+]. Express your answer using two significant figures. [H+][ H + ] = 1.0×10−8 M Previous Answers Correct Since the [OH−] is 100 times greater than the [H+], this expression, [OH−]=100[H+], can be substituted into the ion-product constant expression: Kw==[H+]×100[H+]100[H+]2 Rearranging the equation to solve for [H+]: [H+]===(Kw100)1/2(1.0×10−14100)1/21.0×10−8M Part D Indicate whether the solution is acidic, basic, or neutral.

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Answer:

A: [H⁺] = 2.4 × 10⁻¹¹ M

B: [H⁺] = 1.2 × 10⁻⁶ M

C: [H⁺] =  1.0 × 10⁻⁸ M

D: A and C are basic and B is acid.

Explanation:

Part A: Calculate [H⁺] for [OH⁻] = 4.1×10⁻⁴ M.

We know the ion-product of water Kw is:

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

Then,

[tex][H^{+} ]=\frac{K_{w}}{[OH^{-} ]} =\frac{1.0\times 10^{-14}  }{4.1\times 10^{-4}} =2.4\times10^{-11} M[/tex]

Part B: Calculate [H⁺] for [OH⁻] = 8.5×10⁻⁹ M.

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

[tex][H^{+} ]=\frac{K_{w}}{[OH^{-} ]} =\frac{1.0\times 10^{-14}  }{8.5\times 10^{-9}} =1.2\times10^{-6} M[/tex]

Part C: Calculate [H⁺] for a solution in which [OH⁻] is 100 times greater than [H⁺].

We know that [OH⁻] = 100 . [H⁺]. If we replace this in the ion-product of water:

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

1.0 × 10⁻¹⁴ = [H⁺]. 100. [H⁺]

1.0 × 10⁻¹⁴ = 100 . [H⁺]²

1.0 × 10⁻¹⁶ = [H⁺]²

[H⁺] = √1.0 × 10⁻¹⁶ = 1.0 × 10⁻⁸ M

Part D: Indicate whether the solution is acidic, basic, or neutral.

A solution is acid when [H⁺] > 10⁻⁷ M, basic when [H⁺] < 10⁻⁷ M and neutral when [H⁺] = 10⁻⁷M. So A and C are basic and B is acid.

mahima6824soni

The correct options to the following are:

A. [tex]\rm [H_+] = 2.4 \times 10^-^1^1 M[/tex]

B. [tex]\rm [H^+] = 1.2 \times 10^-^6 M[/tex]

C. [tex]\rm [H^+] = 1.0 \times 10^-^8M[/tex]

D. A and C are basic and B is acid.

What is pH?

The full form of pH is potential hydrogen. pH tells about the acidity and basicity of any liquid.

A. Calculate[tex][H^+] for [OH^-] = 4.1\times10^-^4 M.[/tex]

The ion-product of water Kw is,

Kw = [tex]1.0 \times 10^-^1^4 = [H^+].[OH^-][/tex]

[tex]\rm H^+ = \dfrac{Kw}{[OH^-]} \\\\\\ H^+ =\dfrac{1.0 \times 10^-^1^4 }{[4.1\times10^-^4 M]} = 2.4 \times 10^-^1^1 M[/tex]

B. Calculate[tex][H^+] for [OH^-] = 8.5\times10^-^9 M.[/tex]

Kw = [tex]1.0 \times 10^-^1^4 = [H^+].[OH^-][/tex]

[tex]\rm H^+ = \dfrac{Kw}{[OH^-]} \\\\\\ H^+ =\dfrac{1.0 \times 10^-^1^4 }{[ 8.5 \times 10^-^9]} = 1.2 \times 10^-^6 M[/tex]

C. Calculating the  [H⁺] of a solution in which [OH⁻] is 100 times greater than [H⁺] expression.

Kw = [tex]1.0 \times 10^-^1^4 = [H^+].[OH^-][/tex]

[tex]1.0 \times 10^-^1^4= [H^+]. 100. [H^+]\\\\1.0 \times 10^-^1^6= 100. [H^+]^2\\\\\\[H^+] = \sqrt{1.0 \times 10^-^1^6} = 1.0 \times 10^-^8 M[/tex]

D. Indicate whether the solution is acidic, basic, or neutral.

A solution is acid when [H⁺] > 10⁻⁷ M

Basic when [H⁺] < 10⁻⁷ M and

Neutral when [H⁺] = 10⁻⁷M.

So, A and C are basic and B is acidic.

Thus, A. [tex]\rm [H_+] = 2.4 \times 10^-^1^1 M[/tex]

B. [tex]\rm [H^+] = 1.2 \times 10^-^6 M[/tex]

C. [tex]\rm [H^+] = 1.0 \times 10^-^8M[/tex]

D. A and C are basic and B is acid.

Learn more about ph, here:

https://brainly.com/question/15289741

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