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Chapter 13: Problem 93

What mass of KOH is necessary to prepare \(800.0 \mathrm{mL}\) of a solution having a \(\mathrm{pH}=11.56 ?\)

Short Answer

Expert verified
To prepare 800.0 mL of a solution with a pH of 11.56, 164.6 grams of KOH are necessary.

Step by step solution

01

Calculate the pOH

Since pH + pOH = 14 and we are given the pH, we can calculate pOH as follows: \[ pOH = 14 - pH \] \[ pOH = 14 - 11.56 \] \[ pOH = 2.44 \]
02

Calculate the OH- concentration

Now that we have the pOH value, we can find the OH- concentration using the definition of pOH: \[ pOH = - \log [OH^{-}] \] From this, we can calculate that: \[ [OH^{-}] = 10^{-pOH} \] \[ [OH^{-}] = 10^{-2.44} \] \[ [OH^{-}] = 3.67 \times 10^{-3} mol/L \] **Step 2: Find the moles of KOH**
03

Calculate moles of KOH

In a solution of KOH, there is a one-to-one molar ratio between KOH and OH-ions. Therefore, the concentration of KOH will be the same as the concentration of OH-. We can multiply the concentration of KOH by the volume of the solution to find the moles of KOH: \[ moles\:of\:KOH = concentration \times volume \] \[ moles\:of\:KOH = (3.67 \times 10^{-3}\:mol/L) \times 800.0\:mL \] Note that we must convert the volume in mL to L before doing the calculation: \[ moles\:of\:KOH = (3.67 \times 10^{-3}\:mol/L) \times (800.0/1000\:L) \] \[ moles\:of\:KOH = 2.936\:mol \] **Step 3: Convert moles of KOH to mass**
04

Determine the molar mass of KOH

To convert moles of KOH to mass, we need to know the molar mass of KOH. The molar mass of KOH is: \[ Molar\,mass\,of\,KOH = 39.10\,g/mol\,(K)\,+\,15.99\,g/mol\,(O)\,+\,1.01\,g/mol\,(H) = 56.10\,g/mol \]
05

Calculate the mass of KOH

Now we can find the mass of KOH by multiplying the moles of KOH by its molar mass: \[ mass\:of\:KOH = moles\:of\:KOH \times molar\:mass \] \[ mass\:of\:KOH = 2.936\:mol \times 56.10\:g/mol \] \[ mass\:of\:KOH = 164.6\:g \] Therefore, 164.6 grams of KOH are necessary to prepare 800.0 mL of a solution with a pH of 11.56.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Consider 0.10 \(M\) solutions of the following compounds: \(\mathrm{AlCl}_{3}, \mathrm{NaCN}, \mathrm{KOH}, \mathrm{CsClO}_{4},\) and NaF. Place these solutions in order of increasing \(\mathrm{pH}\).

Rank the following 0.10 \(M\) solutions in order of increasing pH. a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

The pH of a 0.063-M solution of hypobromous acid (HOBr but usually written HBrO) is 4.95. Calculate \(K_{\mathrm{a}}\).

The pOH of a sample of baking soda dissolved in water is 5.74 at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is the solution acidic or basic?
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