Math / Numbers & Operations

QuestionA chemist dissolves $569 . \mathrm{mg}$ of pure barium hydroxide in enough water to make up $120 . \mathrm{mL}$ of solution. Calculate the pH of the solution. (The temperature of the solution is $25^{\circ} \mathrm{C}$ .)
Round your answer to 3 significant decimal places. $\square$ $\square \times 10$ $\square$ ㅁ $\square$ Start over

Studdy Solution

STEP 1

What is this asking? We need to find how acidic or basic a solution is after some barium hydroxide has been dissolved in water, and we'll measure that with pH! Watch out! Barium hydroxide, $Ba(OH)_2$ , provides two hydroxide ions when it dissolves.
Don't forget that!

STEP 2

1. Calculate the moles of barium hydroxide.
2. Calculate the concentration of hydroxide ions.
3. Calculate the pOH.
4. Calculate the pH.

STEP 3

We're given the mass of barium hydroxide, which is 569 mg.
Let's convert this to grams because molar mass is usually given in grams per mole.
Remember, there are 1000 mg in 1 g, so we divide by 1000: $569 \text{ mg} \cdot \frac{1 \text{ g}}{1000 \text{ mg}} = 0.569 \text{ g}$ .

STEP 4

Now, we need the molar mass of $Ba(OH)_2$ .
Looking at the periodic table, we find: Ba = 137.33 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol.
So, the molar mass of $Ba(OH)_2$ is $137.33 + 2 \cdot (16.00 + 1.01) = 171.35 \text{ g/mol}$ .

STEP 5

To find the moles of $Ba(OH)_2$ , we divide the mass by the molar mass: $\frac{0.569 \text{ g}}{171.35 \text{ g/mol}} = 0.00332 \text{ mol}$ .
Awesome!

STEP 6

Remember, each molecule of $Ba(OH)_2$ releases two hydroxide ions ( $OH^-$ ).
So, the moles of $\bf{OH^-}$ are double the moles of $Ba(OH)_2$ : $2 \cdot 0.00332 \text{ mol} = 0.00664 \text{ mol}$ .

STEP 7

The volume of the solution is 120 mL.
We need to convert this to liters: $120 \text{ mL} \cdot \frac{1 \text{ L}}{1000 \text{ mL}} = 0.120 \text{ L}$ .

STEP 8

Now we can calculate the concentration of $\bf{OH^-}$ , which is moles divided by liters: $\frac{0.00664 \text{ mol}}{0.120 \text{ L}} = 0.0553 \text{ M}$ .
This "M" stands for molarity!

STEP 9

The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration: $pOH = -\log_{10}([OH^-])$ .

STEP 10

Let's plug in our value: $pOH = -\log_{10}(0.0553) = 1.257$ .
Looking good!

STEP 11

We know that $pH + pOH = 14$ at $25^{\circ} \mathrm{C}$ .

STEP 12

So, $pH = 14 - pOH = 14 - 1.257 = 12.743$ .
And there we have it!

STEP 13

The pH of the solution is 12.743.

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Studdy Solution

STEP 1

What is this asking? We need to find how acidic or basic a solution is after some barium hydroxide has been dissolved in water, and we'll measure that with pH! Watch out! Barium hydroxide, $Ba(OH)_2$ , provides two hydroxide ions when it dissolves.
Don't forget that!

STEP 2

1. Calculate the moles of barium hydroxide.
2. Calculate the concentration of hydroxide ions.
3. Calculate the pOH.
4. Calculate the pH.

STEP 3

We're given the mass of barium hydroxide, which is 569 mg.
Let's convert this to grams because molar mass is usually given in grams per mole.
Remember, there are 1000 mg in 1 g, so we divide by 1000: $569 \text{ mg} \cdot \frac{1 \text{ g}}{1000 \text{ mg}} = 0.569 \text{ g}$ .

STEP 4

Now, we need the molar mass of $Ba(OH)_2$ .
Looking at the periodic table, we find: Ba = 137.33 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol.
So, the molar mass of $Ba(OH)_2$ is $137.33 + 2 \cdot (16.00 + 1.01) = 171.35 \text{ g/mol}$ .

STEP 5

To find the moles of $Ba(OH)_2$ , we divide the mass by the molar mass: $\frac{0.569 \text{ g}}{171.35 \text{ g/mol}} = 0.00332 \text{ mol}$ .
Awesome!

STEP 6

Remember, each molecule of $Ba(OH)_2$ releases two hydroxide ions ( $OH^-$ ).
So, the moles of $\bf{OH^-}$ are double the moles of $Ba(OH)_2$ : $2 \cdot 0.00332 \text{ mol} = 0.00664 \text{ mol}$ .

STEP 7

The volume of the solution is 120 mL.
We need to convert this to liters: $120 \text{ mL} \cdot \frac{1 \text{ L}}{1000 \text{ mL}} = 0.120 \text{ L}$ .

STEP 8

Now we can calculate the concentration of $\bf{OH^-}$ , which is moles divided by liters: $\frac{0.00664 \text{ mol}}{0.120 \text{ L}} = 0.0553 \text{ M}$ .
This "M" stands for molarity!

STEP 9

The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration: $pOH = -\log_{10}([OH^-])$ .

STEP 10

Let's plug in our value: $pOH = -\log_{10}(0.0553) = 1.257$ .
Looking good!

STEP 11

We know that $pH + pOH = 14$ at $25^{\circ} \mathrm{C}$ .

STEP 12

So, $pH = 14 - pOH = 14 - 1.257 = 12.743$ .
And there we have it!

STEP 13

The pH of the solution is 12.743.

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Studdy Solution

STEP 1

What is this asking? We need to find how acidic or basic a solution is after some barium hydroxide has been dissolved in water, and we'll measure that with pH! Watch out! Barium hydroxide, Ba(OH)2Ba(OH)_2Ba(OH)2​, provides *two* hydroxide ions when it dissolves.Don't forget that!

STEP 2

1. Calculate the moles of barium hydroxide.2. Calculate the concentration of hydroxide ions.3. Calculate the pOH.4. Calculate the pH.

STEP 3

STEP 4

STEP 5

To find the **moles** of Ba(OH)2Ba(OH)_2Ba(OH)2​, we **divide** the mass by the molar mass: 0.569 g171.35 g/mol=0.00332 mol\frac{0.569 \text{ g}}{171.35 \text{ g/mol}} = 0.00332 \text{ mol}171.35 g/mol0.569 g​=0.00332 mol.Awesome!

STEP 6

Remember, each molecule of Ba(OH)2Ba(OH)_2Ba(OH)2​ releases *two* hydroxide ions (OH−OH^-OH−).So, the **moles of** OH−\bf{OH^-}OH− are double the moles of Ba(OH)2Ba(OH)_2Ba(OH)2​: 2⋅0.00332 mol=0.00664 mol2 \cdot 0.00332 \text{ mol} = 0.00664 \text{ mol}2⋅0.00332 mol=0.00664 mol.

STEP 7

The volume of the solution is **120 mL**.We need to **convert** this to liters: 120 mL⋅1 L1000 mL=0.120 L120 \text{ mL} \cdot \frac{1 \text{ L}}{1000 \text{ mL}} = 0.120 \text{ L}120 mL⋅1000 mL1 L​=0.120 L.

STEP 8

Now we can calculate the **concentration** of OH−\bf{OH^-}OH−, which is moles divided by liters: 0.00664 mol0.120 L=0.0553 M\frac{0.00664 \text{ mol}}{0.120 \text{ L}} = 0.0553 \text{ M}0.120 L0.00664 mol​=0.0553 M.This "M" stands for molarity!

STEP 9

The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration: pOH=−log⁡10([OH−])pOH = -\log_{10}([OH^-])pOH=−log10​([OH−]).

STEP 10

Let's plug in our value: pOH=−log⁡10(0.0553)=1.257pOH = -\log_{10}(0.0553) = 1.257pOH=−log10​(0.0553)=1.257.Looking good!

STEP 11

We know that pH+pOH=14pH + pOH = 14pH+pOH=14 at 25∘C25^{\circ} \mathrm{C}25∘C.

STEP 12

So, pH=14−pOH=14−1.257=12.743pH = 14 - pOH = 14 - 1.257 = 12.743pH=14−pOH=14−1.257=12.743.And there we have it!

STEP 13

The pH of the solution is 12.743.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find how acidic or basic a solution is after some barium hydroxide has been dissolved in water, and we'll measure that with pH! Watch out! Barium hydroxide, $Ba(OH)_2$ , provides two hydroxide ions when it dissolves.
Don't forget that!

1. Calculate the moles of barium hydroxide.
2. Calculate the concentration of hydroxide ions.
3. Calculate the pOH.
4. Calculate the pH.

To find the moles of $Ba(OH)_2$ , we divide the mass by the molar mass: $\frac{0.569 \text{ g}}{171.35 \text{ g/mol}} = 0.00332 \text{ mol}$ .
Awesome!

Remember, each molecule of $Ba(OH)_2$ releases two hydroxide ions ( $OH^-$ ).
So, the moles of $\bf{OH^-}$ are double the moles of $Ba(OH)_2$ : $2 \cdot 0.00332 \text{ mol} = 0.00664 \text{ mol}$ .

The volume of the solution is 120 mL.
We need to convert this to liters: $120 \text{ mL} \cdot \frac{1 \text{ L}}{1000 \text{ mL}} = 0.120 \text{ L}$ .

Now we can calculate the concentration of $\bf{OH^-}$ , which is moles divided by liters: $\frac{0.00664 \text{ mol}}{0.120 \text{ L}} = 0.0553 \text{ M}$ .
This "M" stands for molarity!

The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration: $pOH = -\log_{10}([OH^-])$ .

Let's plug in our value: $pOH = -\log_{10}(0.0553) = 1.257$ .
Looking good!

We know that $pH + pOH = 14$ at $25^{\circ} \mathrm{C}$ .

So, $pH = 14 - pOH = 14 - 1.257 = 12.743$ .
And there we have it!