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Chapter 6: Problem 71

What volume of each of the following acids will react completely with \(50.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH} ?\) a. \(0.100 M\) HCl b. \(0.150 M\) HNO \(_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (1 acidic hydrogen)

Short Answer

Expert verified
To react completely with 50.00 mL of 0.200 M NaOH, the required volumes of each acid are as follows: a. 100.00 mL of 0.100 M HCl b. 66.67 mL of 0.150 M HNO₃ c. 50.00 mL of 0.200 M HC₂H₃O₂

Step by step solution

01

Determine the moles of NaOH given

To find the moles of NaOH, we'll use the given volume and concentration of NaOH: Moles of NaOH = (Concentration of NaOH) × (Volume of NaOH) Moles of NaOH = \(0.200 M \times 50.00 mL = 0.200 M \times 50.00 \times 10^{-3} L = 0.0100 mol\)
02

Calculate the moles of required acid

All the reactions given are 1:1 acid-base reactions: One mole of the acid reacts with one mole of the hydroxide ion. Therefore, the moles of the acid required for each reaction are the same as the moles of NaOH. Moles of required acid = Moles of NaOH = 0.0100 mol
03

Calculate volumes of each acid

Now, we'll use the concentrations of the acids and the moles of the acids required for the reaction to calculate the volume of each acid: a. For 0.100 M HCl: Volume = (Moles of HCl) / (Concentration of HCl) Volume = \(0.0100 mol / 0.100 M = 0.100 L = 100.00 mL\) b. For 0.150 M HNO₃: Volume = (Moles of HNO₃) / (Concentration of HNO₃) Volume = \(0.0100 mol / 0.150 M = 0.0667 L = 66.67 mL\) c. For 0.200 M HC₂H₃O₂: Volume = (Moles of HC₂H₃O₂) / (Concentration of HC₂H₃O₂) Volume = \(0.0100 mol / 0.200 M = 0.0500 L = 50.00 mL\)
04

Final answer

To react completely with 50.00 mL of 0.200 M NaOH, the required volumes of each acid are as follows: a. 100.00 mL of 0.100 M HCl b. 66.67 mL of 0.150 M HNO₃ c. 50.00 mL of 0.200 M HC₂H₃O₂

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Most popular questions from this chapter

Describe how you would prepare 2.00 L of each of the following solutions. a. \(0.250 \mathrm{M}\) NaOH from solid \(\mathrm{NaOH}\) b. \(0.250 M\) NaOH from \(1.00 M\) NaOH stock solution c. \(0.100 M K_{2} C r O_{4}\) from solid \(K_{2} C r O_{4}\) d. \(0.100 M K_{2} C r O_{4}\) from \(1.75 M K_{2} C r O_{4}\) stock solution

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472 -g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed 0.2327 g. Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the \(C R C\) Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

Citric acid, which can he ohtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\). A 0.250 -g sample of citric acid dissolved in \(25.0 \mathrm{mL}\) of water requires \(37.2 \mathrm{mL}\) of \(0.105 \mathrm{M}\) NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

A 30.0 -mL. sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is 0.0300 \(M,\) give some possible identities for the unknown base.

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\)
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