Question

Be sure to answer all parts and show all work. A solution is prepared by mixing 18.00 mL of 1.200 M nitric acid (HNO3) with 42.00 mL of 0.400 M potassium hydroxide (KOH). A neutralization reaction occurs between the components in the mixture. a How many moles of nitric acid were originally included in the mixture? b. How many moles of potassium hydroxide were originally included in the mixture? What is the pH of the solution after the neutralization reaction occurs? KOH + HNO3 - > KNO3 + H20

Answer 1

Ta) moles of with acida concentration & volume (litut 1.200 X 0.018 = 0.0216 mol mole of nitur acid into monture = 0.0216 mol te mole of Kolt = j Kolt= cons consentation & volume (litu) 0.400 X 0.042 = 0-0168 mol moles f Kold in the mixture oolol mot KNoz + to Lol Kol & HNO - II 0.0168 0.0216. LE o 4.8x183 mol Since Stoichemetly of the reaction Kol: HNO is moles of koll after reaction 20 moles of two after reaction = 408x107 mos concentration q #woz: 48X6013 = 0.0814 778742JXTOS pHe log (H+) = log (0.08) = 1.0969 PH of th soktion = 1.8966