Question

A Tums tablet weighed 1.319 g. Using the procedure of this experiment, the tablet was dissolved in, and reacted with, an excess amount of HCl (125.00 mL of 0.101 M HCl). The excess unreacted HCl was titrated with NaOH; 18.75 mL of 0.101 M NaOH was used to reach the endpoint. Calculate the moles of HCl that did not react with the antacid. Calculate the moles of HCl that did react with the antacid. Calculate the mass of HCl that did react with the antacid.

Answer 1

Given

Mass of tablet = 1.319 g

Volume of HCl V1 = 125 ml = 0.125 L

Molarity of HCl = 0.101 M (mol/L)

No. of Moles of HCl = Molarity * Volume = 0.125 L * 0.101 mol/L = 0.012625 moles

No. of moles of H+ ions = 0.012625 moles

Volume of NaOH = 18.75 mL = 0.01875 L

Molarity of NaOH = 0.101 M (mol/L)

No. of moles of NaOH = 0.01875 L * 0.101 mol/L = 0.00189375 moles

Each mole of NaOH will give 1 moles of OH-

so no. of moles of OH- = 0.00189375 moles

No. of moles of HCl remaining unreacted = No. of moles of H+ remaining unreacted = No of moles of OH- consumed

No. of moles of HCl remaining unreacted = 0.00189375 moles

No. of moles of acid nuetralized by tablet = Moles of acid present intially - moles of acid remaining unreacted

                                                            = Moles of acid present intially - Moles of OH- used to nuetralize the

                                                                                                                                        remaining acid

                                                           = 0.012645 - 0.00189375 = 0.01075 moles

No. of moles of acid nuetralized by tablet = 0.01075

Answer

(a) No. of moles of HCl remaining unreacted = 0.00189375 moles

(b) No. of moles of HCl reacted with tablet = 0.01075 moles

(c) Mass of Moles of HCl reacted = Molar mass of HCl * No.of moles of HCl reacted = 0.01075 moles * 36.5 g/mol

= 0.392 g