Need a comparison of answers for the following problems... Ihave worked each of them out. Please either a) confirm my answer iscorrect or b) propose an additional answer with solution. Thanks inadvance for your help!
1) Another buffer found in blood is based on the equillibriumbetween dihydrogen phosphate and monohydrogen phosphate. If the pHof a blood sample was 7.10, what would you calculate as the ratioof [H2PO4-] to [HPO4 2-]? The answer I calculated is1.29
2) If you need to perform a reaction in a controlld pHenvironment that was fairly basic, you might choose to use theammonium and amonia buffer system. How many grams of solid ammoniumchloride would you have to add to 2.500 liters of 0.190 M NH3 toobtain a buffered solution of pH= 9.45? The answer I calculated is16.277g NH4Cl
3) Calculate the pH at each of the following points in thetitration of 50.00 mL of a 0.2000 M acetic acid using 0.312 M NaOH(Ka given = 1.8 x 10^-5)
a) initial pH.. The answer I calculated is2.72
b) Halfway to equivelance point... I calculated4.77
c) 5.00 mL before the equivelance point... I calculated5.50
d) At equivelance point... I calculated8.91
e) 5.00 mL beyond equivelance point... I calculated12.26
Need a comparison of answers for the following problems... Ihave worked each of them out. Please either a) confirm my answer iscorrect or b) propose an additional answer with solution. Thanks inadvance for your help!
1) Another buffer found in blood is based on the equillibriumbetween dihydrogen phosphate and monohydrogen phosphate. If the pHof a blood sample was 7.10, what would you calculate as the ratioof [H2PO4-] to [HPO4 2-]? The answer I calculated is1.29
2) If you need to perform a reaction in a controlld pHenvironment that was fairly basic, you might choose to use theammonium and amonia buffer system. How many grams of solid ammoniumchloride would you have to add to 2.500 liters of 0.190 M NH3 toobtain a buffered solution of pH= 9.45? The answer I calculated is16.277g NH4Cl
3) Calculate the pH at each of the following points in thetitration of 50.00 mL of a 0.2000 M acetic acid using 0.312 M NaOH(Ka given = 1.8 x 10^-5)
a) initial pH.. The answer I calculated is2.72
b) Halfway to equivelance point... I calculated4.77
c) 5.00 mL before the equivelance point... I calculated5.50
d) At equivelance point... I calculated8.91
e) 5.00 mL beyond equivelance point... I calculated12.26
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