Q.10.69 - An Introduction to General, Organic, and Biological Chemistry

Question

How many grams of $CaC{O}_3$ are required to neutralize 100mL of stomach acid, which is equivalent to 0.0400 M HCL?

Step-by-Step Solution

Answer

$0.20g$of $CaC{O}_3$ is required.

Step 1: Given Information

the potential of hydrogen (pH) is a scale used to specify the acidity or basicity of a watery solution. Acidic solutions are estimated to have lower pH values than basic or alkaline solutions.

Step 2: Explanation

The reaction is ,

$2\mathrm{HCl}\left(aq\right)+{\mathrm{CaCO}}_3\left(s\right)\to {\mathrm{CaCl}}_2\left(aq\right)+{\mathrm{CO}}_2\left(g\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

Molarity of HCL = $0.040M$

Molar mass of ${\mathrm{CaCO}}_3$= $100.09g$for one mole

$2\mathrm{molHCl}=1\mathrm{mol}{\mathrm{CaCO}}_3$from the equation

now,

$100mL HCL\times \frac{1\mathrm{LHCl}}{1000\mathrm{mLHCl}}\times \frac{0.040\mathrm{molHCl}}{1\mathrm{LHCl}}\times \frac{1\mathrm{mol}{\mathrm{CaCO}}_3}{2\mathrm{molHCl}}\times \frac{100.09\mathrm{g}{\mathrm{CaCO}}_3}{1\mathrm{mol}{\mathrm{CaCO}}_3}$

$=0.2g CaC{O}_3$