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Chapter 16: Problem 31

How much \(\mathrm{NaOH}\) (in grams) is needed to prepare \(546 \mathrm{~mL}\) of solution with a \(\mathrm{pH}\) of 10.00 at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
You need 0.002184 grams of NaOH.

Step by step solution

01

Understanding the pH and pOH

First, we need to understand that the pH of a solution is related to its hydrogen ion concentration. Also, remember that at 25掳C, the sum of pH and pOH is 14. Given that the pH is 10, we can calculate the pOH as follows:\(\text{pOH} = 14 - \text{pH} = 14 - 10 = 4.\)
02

Calculating OH- Concentration

The concentration of hydroxide ions, \([\mathrm{OH}^-]\), in a solution can be found using the formula:\([\mathrm{OH}^-] = 10^{-\text{pOH}}\)Substituting the known value of pOH:\([\mathrm{OH}^-] = 10^{-4} = 0.0001 \mathrm{~mol/L}\)
03

Calculating Moles of NaOH Required

To find how many moles of \(\mathrm{NaOH}\) are needed for the solution, use the formula:\(\text{Moles of NaOH} = [\mathrm{OH}^-] \times \text{Volume of solution in liters}\)Convert the volume from milliliters to liters:\(546 \mathrm{~mL} = 0.546 \mathrm{~L}\)Then calculate the moles:\(\text{Moles of NaOH} = 0.0001 \mathrm{~mol/L} \times 0.546 \mathrm{~L} = 0.0000546 \text{ moles}\)
04

Calculating Mass of NaOH

Calculate the mass of \(\mathrm{NaOH}\) using the formula:\(\text{Mass of NaOH} = \text{Moles of NaOH} \times \text{Molar mass of NaOH}\)The molar mass of \(\mathrm{NaOH}\) is approximately 40.00 \(\mathrm{g/mol}\). So:\(\text{Mass of NaOH} = 0.0000546 \text{ moles} \times 40.00 \mathrm{~g/mol} = 0.002184 \mathrm{~g}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
The pH of a solution indicates its acidity or basicity and is derived from the concentration of hydrogen ions. For any solution, \text{pH}\ is calculated as follows: \( \text{pH} = -\log [\text{H}^+] \). At room temperature (25掳C), the sum of pH and pOH is always 14. This relationship helps us find the concentration of hydroxide ions when given the pH.
  • Given a pH of 10, the pOH is calculated as \(\text{pOH} = 14 - \text{pH} = 4\).
This means the solution has a very low concentration of hydrogen ions but a corresponding relatively high concentration of hydroxide ions.
Hydroxide Ion Concentration
Once the pOH is known, we can calculate the concentration of hydroxide ions, \[ \text{OH}^- \]. The formula used is \( [\text{OH}^-] = 10^{-\text{pOH}} \).
  • For a pOH of 4, \( [\text{OH}^-] = 10^{-4} = 0.0001 \, \text{mol/L} \).
This tells us how many moles of hydroxide ions are present per liter of solution. It's essential for determining the amount of NaOH needed since NaOH dissociates completely to provide one hydroxide ion per molecule.
Molar Mass of NaOH
The molar mass is crucial when converting moles to grams. Sodium hydroxide (NaOH) consists of:
  • Sodium (Na): approximately 23 \, \text{g/mol}
  • Oxygen (O): approximately 16 \, \text{g/mol}
  • Hydrogen (H): approximately 1 \, \text{g/mol}
Adding these together, \( \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \). This simple sum tells us how much one mole of NaOH weighs. When you know the number of moles, multiplying by the molar mass gives the mass in grams.
Solution Concentration
Solution concentration tells us how much solute is present in a given volume of solvent. Here, we need NaOH to make a solution with a specific pH, which determines the hydroxide concentration.
  • We calculated the required moles of NaOH using: \( \text{Moles of NaOH} = [\text{OH}^-] \times \text{Volume in liters} \).
  • For a volume of 546 \, \text{mL} (or 0.546 \, \text{L}), the moles required were \( 0.0000546 \, \text{moles} \).
Finally, by multiplying the moles of NaOH by its molar mass, we calculated the mass needed as 0.002184 \, \text{g}. This ensures the solution has the desired hydroxide concentration, aligning with the pH requirements.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(0.25-M\) aqueous solution of phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right) .\left(K_{\mathrm{a}_{1}}, K_{\mathrm{a}_{2}}\right.\), and \(K_{\mathrm{a}}\), for phosphoric acidare \(7.5 \times 10^{-3}, 6.25 \times 10^{-8},\) and \(4.8 \times 10^{-13},\) respectively.)

A \(20.27-\mathrm{g}\) sample of a metal carbonate \(\left(\mathrm{MCO}_{3}\right)\) is combined with \(500 \mathrm{~mL}\) of a \(1.00 \mathrm{M} \mathrm{HCl}\) solution. The excess \(\mathrm{HCl}\) acid is then neutralized by \(32.80 \mathrm{~mL}\) of \(0.588 \mathrm{M} \mathrm{NaOH}\). Identify \(\underline{\mathrm{M}}\)

Predict the products and tell whether the following reaction will occur to any measurable extent: $$ \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow $$

Calculate the \(\mathrm{pH}\) of an aqueous solution at \(25^{\circ} \mathrm{C}\) that is \(0.095 M\) in hydrocyanic acid \((\mathrm{HCN}) .\left(K_{\mathrm{a}}\right.\) for hydrocyanic acid \(\left.=4.9 \times 10^{-10} .\right)\)

Write equations for the reactions between (a) \(\mathrm{CO}_{2}\) and \(\mathrm{NaOH}(a q),(\mathrm{b}) \mathrm{Na}_{2} \mathrm{O}\) and \(\mathrm{HNO}_{3}(a q)\)
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