Question

ACIDS AND BASES Calculating the pH at equivalence of a titration A chemist titrates 130.0 mL of a 0.6944 M nitric acid (HNO.) solution with 0.8419 M KOH solution at 25 °C. Calculate the pH at equivalence Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH-0 X 5 ?

Answer 1

Ans. Balance reaction - HNO+ KOH KNO3 + H₂O Concentration of HNO 3 = 0.6944 m volume 130 ml - 130 1ooo L = 0.130 L Number of males = Malarity X Valume (in Liture) (n) - 0.6944 m X 0.130 L 3 0.090 male Scanned with CamScanner

Act Equivalence poimut nitric acid will completly meutralize So number of males of HNO3 = number of moles of KOH used by Koh CSScanned with CamScanner

Now KOH con ncentration = 0.8419 m be know malarity Number of males Volume of solution in Litre Scanned with CamScanner

0.8419 m = number of males (n) 7 Litre number of males = 0.8419 mod + 1 x k =0.8419 mol Aut the end of titration number of males of Koh left = 0.8419 mol 0.090 mod 0.7519 mol Now malasrity = Number of males cm) Total volume Scanned with CamScanner

Now malarity =0.751 9 mal 0.7519 mol 12 0.130L til 1.130 L = 0.67 m C Scanned with CamScanner