Question

A sample contains both naoh and nacl. 0.500 g of this sample was dissolved in water to make a 20.0 ml solution and then this solution was titrated by 0.500 mol/l hcl solution. If 22.1 ml of hcl was used to reach the end point, what is the mass % of naoh in the sample?

Answer 1

`Solution:</p><p>moles of HCL =(molarity* volume)/(1000)\\</p><p>=(0.500*22.1)/(1000)\\</p><p>=0.0110\\</p><p>NAOH +HCL -------------->Nacl+H_(2)O\\</p><p>from the reaction NAOH and HCL react with 1:1\\</p><p>so moles of NAOH are needed = 0.0110\\</p><p>(weight)/(molar mass)=0.0110\\</p><p>weight=40*0.0110\\</p><p>      =0.44g\\</p><p>% naoH = (mass of NaoH)/(total mass)* 1000\\</p><p>(0.44*0.5)*100\\</p><p>88%`