Molarity of NaOH-0.103 Molarity of HCl-0.422 Part lIC Scoring Scheme: 3-3-2-1 Us

Question

Molarity of NaOH-0.103

Molarity of HCl-0.422

Part lIC Scoring Scheme: 3-3-2-1 Using the equivalent millimoles HCl/mg sample, the stoichiometry of reaction of HCI with NaOH, and the molar mass of the hydroxide ion, report the gram equivalent OH per gram of antacid for each of your three trials Report them in the same sequence as all your trial data has been reported and use 4 significant figures Vol Vol mmoles mmoles mass tablet(g) antacid(g) HCl(mL)NaOH(mL) HC/mgHCL/tablet antacid 1.304 1.304 1.304 mass Entry # #2: #3: 0.6200 0.6100 0.6120 14.00 14.00 14.00 10.20 9.200 7.800 0.007834 0.008131 0.008340 10.21 10.60 10.87

Explanation / Answer

HCl + NaOH -> NaCl + H2O

As is seen the reaction stoichiometry is 1:1

now, in these titrations amount of OH- of anntacid equals amount of H+ titrated by NaOH.

since these are back titrations. what happens is the antacid neutralizes the H+ but the remaining H+ that is left is neutralized by NaOH

so moles of H+ neutralizsed by anatcid sample ( which is equal to moles OH- it has) is equal to toal HCl - moles of HCl neutralized by NaOH

total n of HCl is = M* vol= 0.422*0.014=0.005908 (molarity *vol in L)

for #1. moles of OH-= total HCl - n of HCl ny NaOH

= 0.005908- (0.103*0.0102)

=0.004857

for #2. moles of OH-= total HCl - n of HCl ny NaOH

= 0.005908- (0.103*0.0092)

= 0.004960

for #3. moles of OH-= total HCl - n of HCl ny NaOH

= 0.005908- (0.103*0.0078)

= 0.005104

answers are in 4 significant figures

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