Question

Chapter 17 .5 4. Calculate the pH a solution made with 25mL of a solution after being titrated with 0.25M HCI in the following amounts a. OmL 12 b. 10mL T c. 38mL 5,04 d. 50mL .40
I have the correct answers written on this questions, but I need help understanding how to get those answers. Thanks!

Answer 1

given moles of NH3-molarity Xvolume 0.38Mx0.025L-0.0095mol a) 10.38M E 0.38-х NH 0.38-x 0.38-x x is very small so that 0.38-x 0.38 →1.80x10-5- 0.38 6.84x10- x-2.62x103 OH-]= x = 2.62 × 10-3 M pOH--log| OH log | 2.62x10% )-2.58 pH 11.42

b) after addition of 10 mL HCl moles of HCI added = 0.25M×0.01L=0.0025mol when 0.0025mol of HCl is added to the above base, same amount of base is converts into conjugate acid then moles of NH30.0095-0.0025 0.007mol moles of conjugate Acid(NH40.0025mol fnad dlume of ba d con ugite Aideual so we can use moles instead fcoscenraicos n Hendeson equation conjugate Acid NH Ka= 1.8×10-5 카K,--log(1.8×10-5)=4.74 hence pOH-4.74+ 0.007 pH 14-4.3 pH - 9.7 0.0025 2

C) after additi on of 38 mL HC moles of HCI = 0.2514× 0.038= 0.0095mol when 0.0095moles of HCl is added, 0.0095mol of NH, converted into Conjugate aci<d Base is completly consumed only Conjugate Acid is present and it undergoes hydrolysis moles of Conjugate acid= 0.0095m Vr -38+25 63.0mL 0.0637 MH1 moles 0.0095mol 0.151M volume 0.063L only Conjugate acidis present and it undergoes hy drolysis 10.151 E 0, 151-X a 0.151-x we know that K^K,-K,-10x10 1.0×10-14 = 5.6×10" a 1.80x10 10 hence 5.6×10-10 since x is very small, (0.151-x)-0.151 →5.6x10-10- x →x= 9.2x10-6M 0.151- x 0.151 H 9.2x10 M pH -log(9.2x10 M) pH -5.04 d) after additi on of 50.0 mL of HCl moles of base(HCI) = 0.25 M× 0.05L = 0.01 25mol when 0.0125moles of HCl is added, 0.0095mol of NH, converted into Conjugate acid and some amount of aci d remains excess HCl remains 0.0125 - 0.00950.003mol Vr =25 + 50 = 75mL=0.075L t 0.003mol 0.075L pHlog (0.04) PH = 1.4