Question

21. 50.0 cm of 1.00 mol dm hydrochloric acid solution, HCl(aq) was added to a piece of pure marble, CaCO3 with a mass of 1.926 g. After all the marble had reacted the resulting solution was put into a 100 cm volumetric flask and the volume made up to the mark with distilled water. 10.0 cm of this solution was then titrated with 1.00 x 10 moldmºsodium hydroxide solution, NaOH(aq).Calculate the volume of the sodium hydroxide solution required to neutralize exactly the excesssipacid present in this 10.0 cm sample during this titration

Answer 1

1 Laul 21) 50 cm3 of 1 mol Ha mass=1.9269 Caloz moles = mass = 1.9269. molar mass Toog mot = 0.01926 mol 50 x103 day? x 1 mol doms After all marble (calog) has reacted » Imoles Caloz react with 2moles HCl < 0.01926 mol 2X0.01926md 50 xlo 3 mol HCl Caloz + 2HCl Cach + H2O + CO2 moes: 0.01926 50xlo mol mol = 0,05 mol mass: 1.926g molar mars! loog/mol 36,5 g/mol moles of HCl reacted = 0.03852 moles! Resulting moles= 0.05-0.03852 mol. Resulting HCl = 0 .01148 ma 7 JHA

Rlow, this unreacted Hel; 0.01148 md. put into a loocm3 flask and dituled Modlarity of resulting solution = moles Volume = (1000 cm3= 12 ) i 100 cm = 0.1L, 0.01148 mol Toox oll M = - HAL 0.1148 moll moldm3 lom of this Hcl titrated with Ixtot Ngoh ( moldm = molto) Using Stardard Molarity equation My = M₂ V₂ He Nooh = 0.1148 x M 10 M lolxV M TV = 11.48 m² Nach Thus, 11.48 cm3 of lach is required to reutralise this excess and is loml.