A 25.00 mL solution of 0.03220 M Na2CO3 is titrated with 0.02510 M Ba(NO3)2 . Calculate pBa2+ following the addition of the given volumes of Ba(NO3)2 . The ?sp for BaCO3 is 5.0×10−9 . 13.90 mLpBa2+= ?epBa2+= 37.40 mLpBa2+=
A 25.00 mL solution of 0.03220 M Na2CO3 is titrated with 0.02510 M Ba(NO3)2 . Calculate pBa2+ following the addition of...
A 25.00 mL solution of 0.03190 M Na2CO3 is titrated with 0.02660 M Ba(NO3)2 . Calculate pBa2+ following the addition of the given volumes of Ba(NO3)2 . The ?sp for BaCO3 is 5.0×10−9 . 15.50 mL pBa2+= ?ep Ba2+= 35.50 mL pBa2+
You titrate 25.00 mL of 0.03290 M Na2CO3 with 0.02530 M Ba(NO3)2. Calculate pBa2 after the following volumes of Ba(NO3)2 are added. Ksp for BaCO3 is 5.0 × 10–9. (a) 12.70 mL (b) Ve (c) 36.30 mL Please show all work and answer all parts
A 25.00 mL solution of 0.03060 M Na,Co, is titrated with 0.02550 M Ba(NO3), Calculate pBa?' following the addition of the given volumes of Ba(NO3), The Kep for Baco, is 5.0 x 10-. 14.90 mL pBa²+ = V. pBa²+ = 37.60 mL pBa2+ =
A 25.00 mL solution of 0.08610 M Nal is titrated with 0.05000 M AgNO . Calculate pAg+ following the addition of the given volumes of AgNO,. The Ksp of Agl is 8.3 x 10-'7. 37.40 mL pAgt = V pAgt = 47.90 mL pAgt =
A 25.00 mL solution of 0.08050 M NaI is titrated with 0.05000 M AgNO3 . Calculate pAg+ following the addition of the given volumes of AgNO3 . The ?sp of AgI is 8.3×10−17 . 35.80 mLpAg+= ?epAg+= 47.90 mLpAg+=
A 25.00 mL solution of 0.08700 M NaI is titrated with 0.05140 M AgNO3 . Calculate pAg+ following the addition of the given volumes of AgNO3 . The
A 0.102 M solution (25.00 mL) of ammonia (pKb =4.76) is titrated with 0.100 M hydrochloric acid. Calculate the pH of the solution after the addition of 15.91 mL acid solution. Present your answer as a numerical value only, to 2 decimal places .
4. (36 pts) Calculate the pH after the addition of 15.00, 25.00, 40.00, and 50.00 mL of 0.400 M NaOH to a solution that contains 100.00 mL of 0.100 M HIO HIO (a)+H20U) »HI(aq)+H, (aq) H 10% (a) H,O)H I0%(a)HO(aq) K, = 2.0 x 10-2 -9 k,-5.0 × 10 a2 4. (36 pts) Calculate the pH after the addition of 15.00, 25.00, 40.00, and 50.00 mL of 0.400 M NaOH to a solution that contains 100.00 mL of 0.100 M...
25 mL of a 0.126 M solution of Ba(OH)2 is titrated with a solution of HCl of unknown molarity. If the equivalence point of the titration is obtained after addition of 28.3 mL of the HCl, the molar concentration of HCl is: Ba(OH)2 + 2 HCl = 2 H2O + BaCl2 A. 0.134 M B. 0.252 M C. 0.285 M D. 0.223 M E. 0.063 M
A 25.00 mL sample of NaOH was titrated with a 0.743 M H2sO, solution. The endpoint of the titration was observed after the addition of 5.00 ml. of H2SO4. Calculate the concentration of the NaOH 2. solution.