Question

What is the pH of a KOH solution that has [H+] = 1.87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5.81 × 10−3 M? What is the pH of a solution of NaCl that has [H+] = 1.00 × 10–7 M?

Answer 1

Answer:

Answers:

1) pH KOH = 12.73

2) pOH KOH = 2.24

3) pH NaCl = 7

Explanation:

pH is the negative logarithm of hydrogen ions concentration (in moles/liter). pH expresses the nature of the solution (acidic, alkaline or neutral).  

In a neutral solution, like pure water, pH = 7 because hydrogen ions (H+) concentration are the same of hydroxide ions (OH-) concentration ([H+] = [OH-] = $1x10^{7} M$). When [H+] are higher than [OH-], then the solution is acidic and when [H+] are smaller than [OH-], the solution is alkaline.  

The formula for calculating pH is $pH = - log [H^{+}]$

1). In the first question we need to calculate pH of a solution of KOH ([H+] = $1.87 x 10^{-13} M$). KOH is a strong base which suffers total ionization in water solution (KOH → K+ + OH-), so character of the solution depends on [OH-] concentration, however the exercise give us [H+] of KOH solution which is related with [OH-] by this expression  

[H^+ ]  ∙[〖OH〗^- ]=1 x 〖10〗^(-14)

then we can use [H+] to calculate pH directly (option 1) or we can calculate [OH-] and then calculate pOH, using $pOH = - log [OH^{-}]$  and   finally, $pH + pOH = 14$ (option 2). I am going to use the first option.

Step 1. Data, question and formula:

            [H+] = $1.87 x 10^{-13} M$

            pH = ? (unknown)

            pH= -log⁡[H^+ ]

Step 2. Replacing the data in the formula

            pH= -log⁡[1.87 x 〖10〗^(-13) ]

     pH = 12.73

Step 3. Analyzing the result

It is recommended to analyze the result to verify that it is logical. In this case we are looking for the pH of an alkaline solution (theoretically pH> 7), the result is 12.73, the pH is greater than 7, so it corresponds to an alkaline substance, in addition the KOH is a strong base, so It is logical that the result is close to 14 which is the maximum value on the pH scale.

2). In the second question we need to calculate pOH of a solution of KOH ([OH-] = $5.81 x 10^{-3} M$). KOH is a strong alkaline solution which suffers total ionization in water (KOH → K+ + OH-), so concentration of each ions is equal to the concentration of KOH, then [KOH] = [K+] = [OH-] = $5.81 x 10^{-3} M$. However, the character of the solution depends only on [OH-] concentration, so it is the only data we need to solve this problem.  

The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.

$pOH = - log [OH^{-}]$

Step 1. Data, question and formula:

            [OH-] = $5.81 x 10^{-3} M$

            pOH = ? (unknown)

           $pOH = - log [OH^{-}]$

Step 2. Replacing the data in the formula

           pOH= -log⁡[5.81 x 〖10〗^(-3) ]

     pOH=2.24

Step 3. Analyzing the result

In this case we are looking for the pOH of an alkaline solution. pOH scale is opposite to pH scale so for a alkaline solution theoretically pOH< 7, the result of this exercise is 2.24, then pOH is lower than 7, so it corresponds to an alkaline substance. It is a logical result for this exercise.

3). Concentration of hydrogen ions in a solution of NaCl is given in the exercise ([H+] = $1 x 10^{-7} M$), then we have to use the following formula:

              $pH = -log [H^{+}]$

Let´s study step by step how to resolve this problem:

Step 1. Data, question and formula:

            [H+] = $1 x 10^{-7}  M$

            pH = ? (unknown)

           $pH = -log [H^{+}]$

Step 2. Replacing the data in the formula

           pH= -log⁡[1 x 〖10〗^(-7) ]

    pH=7,00

Step 3. Analyzing the result

NaCl is an inorganic salt (ionic compound) when it is dissolved in water it suffers total ionization (NaCl → Na+ + Cl-) none of the ions formed will suffer hydrolyzing, it means, none ion will react with water molecule ($H_{2}O$) or its ions (H+ and OH-), so water equilibrium won´t be affected and concentration of water ions will remain equal ([H+] = [OH-] =$1 x 10^{-7} M$), then pH of the NaCl solution is neutral.

So pH of a NaCl solution depends only on pH of water where the salt is dissolved, because salt´s ions don’t react with any other ions or compound. In pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Any aqueous solution in which water equilibrium is not affected ([H+] = [OH-]) is neutral.

Answer 2

The pH of the KOH solution that has {H⁺} = 1.87 × 10⁻¹³ M is 12.728

pH of KOH solution

The pH of the KOH solution that has {H⁺} = 1.87 × 10⁻¹³ M is 12.728

The pH of a solution pH = -㏒[H⁺} where {H⁺} = hydrogen ion concentration.

Given that for out KOH solution, {H⁺} = 1.87 × 10⁻¹³ M,

Substituting this into the equation, we have

pH = -㏒[H⁺}

pH = -㏒[1.87 × 10⁻¹³}

pH = -㏒1.87 - ㏒10⁻¹³

pH = -㏒1.87 - (-13㏒10)

pH = -0.272 + 13

pH = 12.728

So, the pH of the KOH solution that has {H⁺} = 1.87 × 10⁻¹³ M is 12.728

pOH of KOH solution

The pOH of the KOH solution that has {OH⁻} = 5.81 × 10⁻³ M is 2.236

The pOH of a solution pOH = -㏒[OH⁻} where {OH⁻} = hydroxide ion concentration.

Given that for our KOH solution, {OH⁻} = 5.81 × 10⁻³ M,

Substituting this into the equation, we have

pOH = -㏒[OH⁻}

pOH = -㏒[5.81 × 10⁻³}

pOH = -㏒5.81 - ㏒10⁻³

pOH = -㏒5.81 - (-3㏒10)

pOH = -0.764 + 3

pOH = 2.236

So, the pOH of the KOH solution that has {OH⁻} = 5.81 × 10⁻³ M is 2.236

pH of NaCl solution

The pH of the NaCl solution that has {H⁺} = 1.00 × 10⁻⁷ M is 7

The pH of a solution pH = -㏒[H⁺} where {H⁺} = hydrogen ion concentration.

Given that for our NaCl solution, {H⁺} = 1.00 × 10⁻⁷ M,

Substituting this into the equation, we have

pH = -㏒[H⁺}

pH = -㏒[1.00 × 10⁻⁷}

pH = -㏒1.00 - ㏒10⁻⁷

pH = -㏒1.00 - (-7㏒10)

pH = 0 + 7

pH = 7

So, the pH of the NaCl solution that has {H⁺} = 1.00 × 10⁻⁷ M is 7

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