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Chapter 15: Problem 101

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to 500.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15 .\)

Short Answer

Expert verified
To obtain a pH of 2.15 in a 500.0 mL 0.200 M HCl solution, you need to add approximately 17.68 mL of \(1.50 \times 10^{-2}\) M NaOH solution.

Step by step solution

01

Calculate moles of HCl in the initial solution

To find the moles of HCl in the initial solution, we use the formula: moles = Molarity x Volume where Molarity = 0.200 M and Volume = 500.0 mL or 0.500 L (since 1 L = 1000 mL). moles = 0.200 M * 0.500 L = 0.100 moles of HCl Step 2: Calculate the moles of H+ ions after adding NaOH
02

Determine moles of H+ ions required for the desired pH

We use the pH formula to find the moles of H+ ions at a pH of 2.15: pH = -log[H+] 2.15 = -log[H+] By rewriting the equation, we find the concentration of H+ ions: [H+] = 10^{-2.15} = 7.079 x 10^{-3} M Now we calculate the moles of H+ ions in the solution after the neutralization reaction: moles of H+ ions after adding NaOH = (final concentration of H+ ions) x (final volume of solution) The final volume of the solution will be the initial volume of HCl solution (0.500 L) plus the volume of NaOH solution added. Step 3: Calculate the moles of NaOH required
03

Determine moles of NaOH needed to neutralize excess H+ ions

Since NaOH and HCl react in a 1:1 ratio, the moles of NaOH required will be equal to the moles of H+ ions neutralized: moles of H+ ions neutralized = initial moles of HCl - final moles of H+ ions Let x be the volume of NaOH solution to be added: x L * (1.50 x 10^{-2} M) = 0.100 moles HCl - [(7.079 x 10^{-3} M) * (0.500 L + x)] Step 4: Solve for the volume of NaOH solution
04

Calculate the volume of NaOH solution

Solve the equation above for the volume (x) of NaOH solution required: x L * (1.50 x 10^{-2} M) = 0.100 - [(7.079 x 10^{-3} M) * (0.500 + x)] Now, let's isolate x on one side of the equation: x (1.50 x 10^{-2} M + 7.079 x 10^{-3} M) = 0.100 - 7.079 x 10^{-3} M * 0.500 Divide both sides by the coefficients of x: x = (0.100 - [7.079 x 10^{-3} * 0.500])/(1.50 × 10^{-2} M + 7.079 x 10^{-3} M) Now we can calculate the value of x: x ≈ 0.01768 L To obtain the volume in milliliters, multiply this value by 1000: 0.01768 L * 1000 = 17.68 mL So, we need to add approximately 17.68 mL of 1.50 x 10^{-2} M NaOH to 500.0 mL of 0.200 M HCl solution to obtain a pH of 2.15.

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Most popular questions from this chapter

Consider a solution formed by mixing 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) HOCl, 25.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) \(\mathrm{NaOH}, 25.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2},\) and 10.0 \(\mathrm{mL}\) of 0.150 \(M \mathrm{KOH}\) . Calculate the \(\mathrm{pH}\) of this solution.

A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular fluid at \(\mathrm{pH}=7.15 ?\) $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$ b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

Consider the titration of 150.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HI}\( by 0.250 \)\mathrm{M}\( \)\mathrm{NaOH}$ . a. Calculate the pH after 20.0 \(\mathrm{mL}\) of NaOH has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=7.00 ?\)

Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 \(\mathrm{mL}\) of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{MNH}_{4} \mathrm{Cl}} \\ {\text { b. } 0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}}\end{array} $$ Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?
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