Question

calculating the ph of a weak base titrated with a strong acid

An analytical chemist is titrating 153.8 ml of a 0.5900 M solution of trimethylamine (CH), N) with a 0.3700 M solution of HNOJ. The pl, of trimethylamine is 4.19. Calculate the pH of the base solution after the chemist has added 34.2 mL of the HNO, solution to it Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. pH = 0 XS ? 888

Answer 1

Answer:-

This question is solved by using the simple concept of calculation of pH of a buffer solution using simple concept of Henderson Hessalbach equation.

The answer is given in the image,

Enjዚህን :- {€13731]=0.5900m, 1538mm CHAD ] = 0-3700 M, 34.2 m2 -pko = 4.19 pkee 14-pky = 14 -4.19 -pka=9.81 (CH3)₂N + H+ (CH3)NH so_Moles (CH3)4=05JOOX 153.8 Looo = 0.090742 mol. Moles q HMOz added = 0.3700X 34.2 1000 C1322 MH =0.012654 mol Netmoles of (CH3)2 Mremaines -0.090742-0.012654 20.078088mole pH = pka + log (CH32 M) 6CC4133444+7 pH=9.81 + log 0.078088 0.012654 PH = 9.81 +0.79