(d) If 42.7 mL of 0.208 M HCl solution is needed to

Problem 4.82dSolution Stoichiometry and Chemical Analysis (Section)(a) How many millimeters of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101M Ba(OH) sol2ion(b) How many millimeters of 0.125 M H SO are nee2d to4eutralize 0.200 g of NaOH(c) If 55.8 mL of a BaCl soluti2n is needed to precipitate all the sulfate ion in a 752-mg sampleof Na SO , what is the molarity of the BaCl solution 2 4 2(d) If 42.7 mL of 0.208 M HCl solution is needed to neutralize a solution of Ca(OH) , how 2many grams of Ca(OH) must be in the solution 2 Step-by-step solution Step 1 of 7 ^(a)To calculate the volume of HCl needed to neutralize Ba(OH) solution. 2 M V 1 1 M V 2 2 = n 1 n 2Here, M , V , and n are molarity, volume, and number of moles of Ba(OH) ; where M , V , and 1 1 1 2 2 2n2efer to HCI.The balanced chemical between is as follows:2 HCl + Ba(OH) BaC2 + 2H O 2 2From the balanced equation, 2 moles of HCl is needed to neutralize 1 mole of Ba(OH) . 2 0.101M x 50.0 mL 0.120M x V 2 = 1 2 2 x 0.101M x 50.0 mL V2= 0.120M V2= 83.3 mLTherefore, volume of HCl will be .\n