After treating an antacid with 5.00 ml of .1065 M NaOH, the untreated acid was d
ID: 608780 • Letter: A
Question
After treating an antacid with 5.00 ml of .1065 M NaOH, the untreated acid was diluted to exactly 250 mL with distilled water in a volumetric flask. 25.00 mL of that diluted acid required 17.03 mL of the same NaOH solution to neutralize it. Calculate the mole of acid neutralized by the tablet.Explanation / Answer
well this is chemistry isnt it not mathematics but i can suggest the basic idea first step is writing the chemical equation and balance it , in this case it seems to be pretty simple NaOH + HcL ------------> NacL + H2O so basically you need one mole of NaOH to neutralise one mole of HcL .1M means i assume .1 moles per litre so how many moles of NaOH do you have here 5/1000 * .1 = 5 * 10^-4 moles of NaOH so this can only neutralise 5 * 10^-4 moles of Hcl according to our equation now maybe you forgot the concentration of HCl , is it .1M also or 1M ? if it is .1M then you have .1 moles in 1000 ml so 20 ml will have 20/1000 * .1 = 20 * 10^-4 moles so 1 gram of the antacid neutralised 15 * 10^-4 moles of HcL and left 5 * 10^-4 moles for the NaOh to neutralise so this is the answer if the 20 mL was .1M concentration , but i think it maybe 1M usually for acid, please check and then do accordingly