1. If you are to make 1 L of a 0.1 M H2PO4- / HPO42- (pKa2 = 6.9) buffer at pH 6.9 and you start with the NaH2PO4, how many grams of NaH2PO4 is needed? Do you then add HCl or NaOH to make the final buffer and how moles of HCl or NaOH is needed? MW for NaH2PO4 is 120.
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
To make 1.00 L of a 0.100 M H2PO4- / HPO42- buffer at pH 6.90 and you start with the NaH2PO4, A. Calculate the mass of NaH2PO4 needed to make the buffer. B. Should HCl or NaOH be added to make the final buffer? Calculate the number of moles of HCl or NaOH that needs to be added. Note: formula mass of NaH2PO4 is 120.
You are instructed to create 400. mL of a 0.40 M
phosphate buffer with a pH of 6.4. You have phosphoric acid and the
sodium salts NaH2PO4,
Na2HPO4, and Na3PO4
available. (Enter all numerical answers to three significant
figures.)
H3PO4(s) +
H2O(l)
H3O+(aq) +
H2PO4−(aq)
Ka1 = 6.9 ✕ 10−3
H2PO4−(aq) +
H2O(l)
H3O+(aq) +
HPO42−(aq)
Ka2 = 6.2 ✕ 10−8
HPO42−(aq) +
H2O(l)
H3O+(aq) +
PO43−(aq)
Ka3 = 4.8 ✕ 10−13
Which of the available chemicals will you use...
You are instructed to create 200. mL of a 0.63 M phosphate buffer with a pH of 6.0. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s) + H2O(l) equilibrium reaction arrow H3O+(aq) + H2PO4−(aq) Ka1 = 6.9 ✕ 10−3 H2PO4−(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + HPO42−(aq) Ka2 = 6.2 ✕ 10−8 HPO42−(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + PO43−(aq) Ka3 = 4.8 ✕ ...
You are instructed to create 400. mL of a 0.39 M phosphate buffer with a pH of 6.2. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s) + H2O(l) = H30+ (aq) + H2P04 (aq) Kai = 6.9 x 10-3 H2PO4 (aq) + H20(1) =H30+ (aq) + HPO42-(aq) Ka2 = 6.2 x 10-8 HPO42-(aq) + H20(I) = H30+(aq) + PO43-(aq) Ka3 = 4.8 x 10-13 Which of...
A buffer is to be made up with a mixture of H2PO4- and HPO42-, how many grams of HPO42- is needed for a buffer of pH=7.5 if the solution contains .0015 mol of H2PO4- Ka of H2PO4- is 6.2x10-8
Part 1: Which pair of substances would make a good buffer system? a. H2PO4-/HPO42- b. H2PO4-/HPO42 c. SO42-/H2SO4 d. HCl/NaOH Part 2: Should the concentrations of the two buffer components be present at relatively high and concentrations close to each other or low with concentrations far from each other ? a. far and low b. close and high
You are instructed to create 400. mL of a 0.39 M phosphate buffer with a pH of 6.2. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s) + H20(I) =H30+ (aq) + H2PO4 (aq) Kai = 6.9x10-3 H2PO4 (aq) + H20(1) =H30+ (aq) + HPO42-(aq) Ka2 = 6.2 x 10-8 HPO42-(aq) + H20(1) = H30+(aq) + PO43-(aq) Ka3 = 4.8x10-13 Which of the available chemicals will you...
You are instructed to create 400. mL of a 0.39 M phosphate buffer with a pH of 6.2. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s) + H20(1) =H30+ (aq) + H2PO4 (aq) Kai = 6.9x10-3 H2PO4 (aq) + H20() = H30+(aq) + HPO42- (aq) Ka2 = 6.2x 10-8 HPO42-(aq) + H20(I) = H30+(aq) + PO43-(aq) Ka3 = 4.8 x 10-13 Which of the available chemicals...
You are instrucbed to create 900. mL of a 0.59 M phosphate buffer with a pH of 6.4. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. numerical answers to three significant H3PO4(s)H20)H3 aH2PO4(aq) H2PO4-(aq) + H2O(りーH3O + (aq) + HP042-(aq) Ka2-6.2×10-8 HPO()H20()H3)PO43 (aq) Ka 4.8x10-13 Ka 6.9x10-2 Which of the available chemicals will you use for the acid component of your buffer? O H3PO4 O Na2HPO Na3PO4 Which of the available chemicals will you use...
In our experiment, we will be using a portion of the phosphate buffer system that is based upon the following equilibrium: H2PO4- HPO42- + H+ pKa = 7.2 In this case, H2PO4- will act as the acid and HPO42- will act as the base. Materials: 1M NaOH: 40.01 g/L of solution 1M HCl: 83 mL conc. HCl/L of solution Potassium phosphate, dibasic, K2HPO4, MW= 174.18 Potassium phosphate, monobasic, KH2PO4 MW= 136.09 **I already preformed this lab, but I struggled a...