Math  /  Numbers & Operations

QuestionA chemist dissolves 863.mg863 . \mathrm{mg} of pure sodium hydroxide in enough water to make up 180.mL180 . \mathrm{mL} of solution. Calculate the pH of the solution. (The temperature the solution is 25C25^{\circ} \mathrm{C}.)
Round your answer to 3 significant decimal places. \square ×10\times 10

Studdy Solution

STEP 1

What is this asking? We need to find how acidic or basic a solution is after we dissolve some sodium hydroxide in water, and give the answer using the pH scale. Watch out! Remember that sodium hydroxide is a strong base, so it completely dissociates in water.
Don't forget those conversions between milligrams, grams, and liters!
Also, remember the relationship between pOH and pH.

STEP 2

1. Calculate the moles of sodium hydroxide.
2. Calculate the molarity of hydroxide ions.
3. Calculate the pOH of the solution.
4. Calculate the pH of the solution.

STEP 3

First, we **convert** the mass of sodium hydroxide from milligrams to grams.
We know that 1 g=1000 mg1 \text{ g} = 1000 \text{ mg}, so we have 863 mg1000 mg/g=0.863 g\frac{863 \text{ mg}}{1000 \text{ mg/g}} = 0.863 \text{ g} of sodium hydroxide.
Why grams?
Because molar mass is in grams!

STEP 4

Next, we **calculate** the number of moles of sodium hydroxide.
The molar mass of sodium hydroxide (NaOH) is approximately 23+16+1=40 g/mol23 + 16 + 1 = 40 \text{ g/mol}.
So, we have 0.863 g40 g/mol=0.021575 mol\frac{0.863 \text{ g}}{40 \text{ g/mol}} = \textbf{0.021575 mol} of sodium hydroxide.
Awesome!

STEP 5

Now, we need to **convert** the volume of the solution from milliliters to liters.
Since 1 L=1000 mL1 \text{ L} = 1000 \text{ mL}, we have 180 mL1000 mL/L=0.180 L\frac{180 \text{ mL}}{1000 \text{ mL/L}} = 0.180 \text{ L}.
Liters are needed for molarity!

STEP 6

Since sodium hydroxide is a strong base, it completely dissociates in water, meaning that the concentration of hydroxide ions (OHOH^-) is equal to the concentration of sodium hydroxide.
So, the molarity of hydroxide ions is 0.021575 mol0.180 L=0.11986 M\frac{\textbf{0.021575 mol}}{0.180 \text{ L}} = \textbf{0.11986 M}.
Almost there!

STEP 7

We can **calculate** the pOH using the formula pOH=log10[OH]pOH = -\log_{10}[OH^-], where [OH][OH^-] is the concentration of hydroxide ions.
Plugging in our value, we get pOH=log10(0.11986)0.921pOH = -\log_{10}(\textbf{0.11986}) \approx \textbf{0.921}.
Getting close!

STEP 8

Finally, we can **calculate** the pH using the relationship pH+pOH=14pH + pOH = 14 at 25C25^{\circ} \mathrm{C}.
So, pH=140.921=13.079pH = 14 - \textbf{0.921} = \textbf{13.079}.
We did it!

STEP 9

The pH of the solution is **13.079**.

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