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Chapter 4: Problem 86

An \(8.65-g\) sample of an unknown group 2 metal hydroxide is dissolved in $85.0 \mathrm{~mL}$ of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M}\) \(\mathrm{HCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+},\) or $\mathrm{Ba}^{2+} ?$

Short Answer

Expert verified
The molar mass of the unknown metal hydroxide is \(121.64\,g/mol\), and its metal cation is \(Sr^{2+}\), with the metal hydroxide being \(Sr(OH)_2\).

Step by step solution

01

Calculate the moles of HCl used in titration

: We use the molarity and volume of the HCl solution to find the moles: Molarity (M) = moles of solute/liters of solution Moles of HCl = Molarity × Volume The given volume of HCl is 56.9 mL. We convert it to liters by dividing it by 1000: \(56.9\, mL \times \frac{1\,L}{1000\,mL} = 0.0569\, L\) Moles of HCl = 2.50 M × 0.0569 L = 0.14225 moles
02

Determine the moles of the metal hydroxide

: The balanced equation for the reaction between the unknown metal hydroxide (M(OH)2) and HCl is: \(M(OH)_{2}(aq) + 2\,HCl(aq) \rightarrow MCl_{2}(aq) + 2\,H_{2}O(l)\) From the balanced equation, we see that 1 mole of metal hydroxide reacts with 2 moles of HCl. We can use the stoichiometric ratios to find moles of the metal hydroxide: Moles of M(OH)2 = (Moles of HCl) / 2 Moles of M(OH)2 = 0.14225 moles / 2 = 0.071125 moles
03

Calculate the molar mass of the metal hydroxide

: We have the mass and moles of the metal hydroxide, so we can find its molar mass: Molar mass = mass / moles Molar mass = 8.65 g / 0.071125 moles = 121.64 g/mol
04

Identify the metal cation

: The molar mass of the metal hydroxide is 121.64 g/mol. We can identify the metal cation by comparing with the known molar mass of Group 2 metal hydroxides: Ca(OH)2: 40.08 (Ca) + 2(15.999 (O) + 1.00784 (H)) = 74.10 g/mol Sr(OH)2: 87.62 (Sr) + 2(15.999 (O) + 1.00784 (H)) = 121.64 g/mol Ba(OH)2: 137.33 (Ba) + 2(15.999 (O) + 1.00784 (H)) = 171.35 g/mol Based on the molar mass, the metal hydroxide is Sr(OH)2 and hence, the metal cation is Sr²⁺.

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Most popular questions from this chapter

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of $0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}$ solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If $55.8 \mathrm{~mL}\( of a \)\mathrm{BaCl}_{2}$ solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of $\mathrm{Na}_{2} \mathrm{SO}_{4}\(, what is the molarity of the \)\mathrm{BaCl}_{2}$ solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{MHCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make $1000.0 \mathrm{~L}\( of a \)0.150 \mathrm{M}$ aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2},\) or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} . \mathrm{A}\) friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) Silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

Uranium hexafluoride, \(\mathrm{UF}_{6},\) is processed to produce fuel for nuclear reactors and nuclear weapons. UF \(_{6}\) can be produced in a two-step reaction. Solid uranium (IV) oxide, \(\mathrm{UO}_{2}\), is first made to react with hydrofluoric acid (HF) solution to form solid UF \(_{4}\) with water as a by-product. \(U F_{4}\) further reacts with fluorine gas to form UF \(_{6}\). (a) Write the balanced molecular equations for the conversion of \(U O_{2}\) into \(U F_{4}\) and the conversion of \(U F_{4}\) to \(U F_{6}\) (b) Which step is an acid-base reaction? (c) Which step is a redox reaction?
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