Explanation Given: Mass of Ca ( OH ) 2 is 0 .300 g . Concentration of HNO 3 is 1 .40 M . Volume of HNO 3 is 50 .0 mL ( 0 .050 L ) . Volume of the solution is 75 mL ( 0 .075 L ) . The balanced chemical equation corresponding to the reaction that takes place is, Ca ( OH ) 2 ( a q ) + 2HNO 3 ( a q ) → Ca ( NO 3 ) 2 ( a q ) + 2H 2 O ( a q ) The above reaction is a neutralization reaction. Formula used: M = n V Where, M is the molarity. n is the number of moles. V is the volume of the solution. Calculation: The molar mass of Ca ( OH ) 2 is, = ( 40 g/mol ) + ( 2×16 g/mol ) + ( 2×1 g/mol ) = 74 g/mol Number of moles calculated by the formula, n = m M Where, n is the number of moles. m is the mass of the molecule. M is the molar mass of the molecule. Substitute the value of mass of Ca ( OH ) 2 and molar mass of Ca ( OH ) 2 in the above formula. n = 0 .3 g 74 g/mol = 0 .004 mol Molarity is calculated by the formula, M = n V ....... (1) Where, M is the molarity. n is the number of moles. V is the volume of the solution. Rearrange the above equation to calculate the number of moles for HNO 3 . n = M × V Substitute the value of molarity of HNO 3 and volume of the solution in the above expression. n = 0 .050 L × 1 .40 M = 0 .07 moles According to the stated reaction, 1 mol of Ca ( OH ) 2 reacts with 2 moles of HNO 3 . One mole of Ca ( OH ) 2 reacts with moles of HNO 3 = 2

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Chapter 2, Problem 76E

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To determine: The value of the pH of the final solution.

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Chapter 2, Problem 75E

Chapter 2, Problem 77E

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