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What volume of \(0.125 \mathrm{M}\) oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) is required to react with \(35.2 \mathrm{mL}\) of \(0.546 \mathrm{M} \mathrm{NaOH} ?\) $$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$

Short Answer

Expert verified
76.88 mL of 0.125 M oxalic acid is required.

Step by step solution

01

Understand the Reaction Stoichiometry

The balanced chemical equation for the reaction between oxalic acid and sodium hydroxide is given by:\[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\]From the equation, we see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.
02

Calculate Moles of NaOH

First, we calculate the number of moles of \(\mathrm{NaOH}\) using its concentration and volume:\[\text{Moles of } \mathrm{NaOH} = \text{Concentration} \times \text{Volume in Liters} = 0.546 \, \mathrm{M} \times 0.0352 \, \mathrm{L} = 0.0192192 \, \mathrm{mol}\]
03

Determine the Required Moles of Oxalic Acid

Using the stoichiometry of the reaction, \(1\) mole of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) reacts with \(2\) moles of \(\mathrm{NaOH}\). Therefore, the moles of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) required are:\[\text{Moles of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} = \frac{0.0192192 \, \mathrm{mol}}{2} = 0.0096096 \, \mathrm{mol}\]
04

Calculate the Volume of Oxalic Acid Solution Needed

Finally, calculate the volume of \(0.125 \, \mathrm{M}\) oxalic acid needed to provide \(0.0096096 \, \mathrm{mol}\):\[\text{Volume in Liters} = \frac{\text{Moles of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}}{\text{Concentration of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}} = \frac{0.0096096 \, \mathrm{mol}}{0.125 \, \mathrm{M}} = 0.0768768 \, \mathrm{L}\]To convert this volume to milliliters:\[0.0768768 \, \mathrm{L} \times 1000 = 76.88 \, \mathrm{mL}\]

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Key Concepts

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

Reaction Stoichiometry
Understanding reaction stoichiometry is crucial in any chemical reaction. It tells us the exact ratio in which different reactants combine to form products. In our specific problem, the balanced chemical equation is given as: \[ \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \] The equation indicates that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide. This stoichiometric relationship is key. It guides us in determining how much of one reactant is necessary to completely react with a given quantity of another. In our case, the 1:2 ratio signals that for every mole of oxalic acid, two moles of NaOH are needed. This foundational step ensures that calculations about reactants and products adhere to the laws of conservation of mass.
Moles Calculation
The next step involves calculating the moles of sodium hydroxide (NaOH) present in the given solution. Moles are a unit of measurement which quantify the amount of a substance. To find the moles, we employ the formula: \[ \text{Moles} = \text{Concentration} \times \text{Volume in Liters} \] In our exercise, the concentration provided is 0.546 M (moles per liter), and the volume is 35.2 mL. Before using the formula, it's important to convert milliliters to liters. Remember:- 1 L = 1000 mL - Therefore, 35.2 mL = 0.0352 L With these conversions in place, you can determine the moles of NaOH:\[ 0.546 \text{ M} \times 0.0352 \text{ L} = 0.0192192 \text{ mol} \] Thus, there are approximately 0.0192 moles of NaOH in the solution.
Solution Concentration
Solution concentration is a measure of how much solute is dissolved in a given volume of solvent. Usually expressed in molarity (M), concentration plays a vital role in calculating the volume required for reactions. In our problem, the concentration of oxalic acid (\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)) is given as 0.125 M. This means there are 0.125 moles of oxalic acid in every liter of its solution.Using our stoichiometry knowledge, we found earlier that 0.0096096 mol of oxalic acid is necessary to fully react with our sodium hydroxide. Therefore, the concentration helps us determine how much solution we need to supply this exact mole quantity.
Volume Conversion
Converting volumes between units is frequently required, especially in chemistry where solutions are often measured in liters or milliliters. Our task involves converting the final solution volume from liters back to milliliters for practicality and comprehension.In the last calculation step, you find the volume of the oxalic acid solution necessary using the relation:\[ \text{Volume in Liters} = \frac{\text{Moles of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}}{\text{Concentration of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}} = \frac{0.0096096 \text{ mol}}{0.125 \text{ M}} = 0.0768768 \text{ L} \]To convert this volume into milliliters:- Multiply the volume in liters by 1000 (since 1 L = 1000 mL)- Hence, 0.0768768 L becomes 76.88 mLThis conversion ensures the volume is presented in a unit more commonly used in laboratory settings, making it intuitively easier to measure and use.

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Most popular questions from this chapter

Phosphate in urine can be determined by spectrophotometry. After removing protein from the sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate can be measured at \(650 \mathrm{nm}\) and is directly related to the urine phosphate concentration. A 24 -hour urine sample was collected from a patient; the volume of urine was 1122 mL. The phosphate in a 1.00 mL portion of the urine sample was converted to the blue polymolybdate and diluted to 50.00 mL. A calibration curve was prepared using phosphate-containing solutions. (Concentrations are reported in grams of phosphorus (P) per liter of solution.) $$\begin{array}{|c|c|}\hline \text { Solution (mass P/L) } & \begin{array}{c}\text { Absorbance at } 650 \mathrm{nm} \\\\\text { in a } 1.0-\mathrm{cm} \text { cell }\end{array} \\\\\hline 1.00 \times 10^{-6} \mathrm{g} & 0.230 \\\\\hline 2.00 \times 10^{-6} \mathrm{g} & 0.436 \\\\\hline 3.00 \times 10^{-6} \mathrm{g} & 0.638 \\\\\hline 4.00 \times 10^{-6} \mathrm{g} & 0.848 \\ \hline \text { Urine sample } & 0.518 \\\\\hline\end{array}$$ (a) What are the slope and intercept of the calibration curve? (b) What is the mass of phosphorus per liter of urine? (c) What mass of phosphate did the patient excrete in the one-day period?

Anhydrous calcium chloride is a good drying agent because it will rapidly pick up water. Suppose you have stored some carefully dried \(\mathrm{CaCl}_{2}\) in a desiccator. Unfortunately, someone did not close the top of the desiccator tightly, and the \(\mathrm{CaCl}_{2}\) became partially hydrated. A \(150-\mathrm{g}\) sample of this partially hydrated material was dissolved in \(80 \mathrm{g}\) of hot water. When the solution was cooled to \(20^{\circ} \mathrm{C}, 74.9 \mathrm{g}\) of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) precipitated. Knowing the solubility of calcium chloride in water at \(20^{\circ} \mathrm{C}\) is \(74.5 \mathrm{g} \mathrm{CaCl}_{2} / 100 \mathrm{g}\) water, determine the water content of the 150 -g sample of partially hydrated calcium chloride (in moles of water per mole of \(\mathrm{CaCl}_{2}\) ).

Commercial sodium "hydrosulfite" is \(90.1 \%\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} .\) The sequence of reactions used to prepare the compound is $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})$$ $$\mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})$$ (a) What mass of pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) can be prepared from \(125 \mathrm{kg}\) of \(\mathrm{Zn}, 500 .\) g of \(\mathrm{SO}_{2},\) and an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (b) What mass of the commercial product would contain the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) produced using the amounts of reactants in part (a)?

A mixture of butene, \(\mathrm{C}_{4} \mathrm{H}_{8},\) and butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\) is burned in air to give \(\mathrm{CO}_{2}\) and water. Suppose you burn 2.86 g of the mixture and obtain \(8.80 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(4.14 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What are the mass percentages of butene and butane in the mixture?

Boron forms a series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\). $$\mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\text { excess } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{\gamma}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ If \(0.148 \mathrm{g}\) of one of these compounds gives \(0.422 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2},\) what is its empirical formula?

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