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Chapter 4: Q78E (page 224)

What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-MCa(OH)2 solution?

Ca(OH)2(aq) + 2HBr(aq)⟶CaBr2(aq) + 2H2 O(l)

Short Answer

Expert verified

238 mL of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-M Ca(OH)2 solution

Step by step solution

01

Given data 

This titration is between acid and base. For this formula we use

\({M_a}{V_a} = {M_b}{V_b}\)

where a = acid (HBr) and b = base (Ca(OH)2)

Molarity of HBr = 0.0105 M and Volume of HBr =??

Molarity of Ca(OH)2 = 0.0100 M Volume of Ca(OH)2 = 125 mL

02

Determine volume of HBr

Ca(OH)2(aq) + 2HBr(aq)⟶CaBr2(aq) + 2H2 O(l)

\({M_a}{V_a} = {M_b}{V_b}\)

By putting the values in the above equation,

\(\begin{aligned}{}1 \times 0.0105M \times X = 0.0100\,M \times 2 \times 125\,mL\\X = \frac{{0.0100\,M \times 2 \times 125\,mL}}{{0.0105M}}\\X = 238.09\,mL\end{aligned}\)

238 mL of HBr

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Most popular questions from this chapter

Balance the following equations:

\(\begin{array}{l}\left( a \right)\,Ag\left( s \right) + {H_2}S\left( g \right) + {O_2}\left( g \right) \to A{g_2}S\left( s \right) + {H_2}O\left( l \right)\\\left( b \right)\,{P_4}\left( s \right) + {O_2}\left( g \right) \to {P_4}{O_{10}}\left( s \right)\\\left( c \right)\,Pb\left( s \right) + {H_2}O\left( l \right) + {O_2}\left( g \right) \to Pb{\left( {OH} \right)_2}\left( s \right)\\\left( d \right)\,Fe\left( s \right) + {H_2}O\left( l \right) \to F{e_3}{O_4}\left( s \right) + {H_2}\left( g \right)\\\left( e \right)\,S{c_2}{O_3}\left( s \right) + S{O_3}\left( g \right) \to S{c_2}{\left( {S{O_4}} \right)_3}\left( s \right)\\\left( f \right)\,C{a_3}{\left( {P{O_4}} \right)_2}\left( {aq} \right) + {H_3}P{O_4}\left( {aq} \right) \to Ca{\left( {{H_2}P{O_4}} \right)_2}\left( {aq} \right)\\\left( g \right)\,Al\left( s \right) + {H_2}S{O_4}\left( {aq} \right) \to A{l_2}{\left( {S{O_4}} \right)_3}\left( s \right) + {H_2}\left( g \right)\\\left( h \right)\,TiC{l_4}\left( s \right) + {H_2}O\left( g \right) \to Ti{O_2}\left( s \right) + HCl\left( g \right)\end{array}\)

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