A sample of P2O5 contains some H3PO4 impurity. A 0.405-g sample is reacted with water (P2O5...
A sample of P2O5 contains some H3PO4 impurity. A
0.405-g sample is reacted with
water (P2O5 + 3H2O → 2H3PO4), and the resulting solution is
titrated with 0.250 M NaOH(H3PO4 → Na2HPO4). If 42.5 mL is required
for the titration, what is the percent of H3PO4 impurity?
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A sample of P2O5 contains some H3PO4 impurity. A
0.405-g sample is reacted with
water (P2O5...
A 31.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH...
A 31.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 26.08 mL of NaOH solution is added. What is the concentration of the unknown H3PO4 solution? The neutralization reaction is H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)
A 27.00 mL sample of an unknown H3PO4 solution is titrated with a 0.110 M NaOH...
A 27.00 mL sample of an unknown H3PO4 solution is titrated with a 0.110 M NaOH solution. The equivalence point is reached when 28.08 mL of NaOH solution is added. What is the concentration of the unknown H3PO4 solution? The neutralization reaction is H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)
A 26.00 mL sample of an unknown H3PO4 solution is titrated with a 0.120 M NaOH...
A 26.00 mL sample of an unknown H3PO4 solution is titrated with a 0.120 M NaOH solution. The equivalence point is reached when 25.38 mL of NaOH solution is added.What is the concentration of the unknown H3PO4 solution? The neutralization reaction is H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)
If a 16.5 mL sample of 1.4 M solution of each of the following acids is reacted with 0.90 M NaOH, how many milliliters...
If a 16.5 mL sample of 1.4 M solution of each of the following acids is reacted with 0.90 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point? (a) 16.5 mL of H,So, titrated with 0.90 M NaOH volume of NaOH mL total volume mL (b) 16.5 mL of HCl titrated with 0.90 M NaOH volume of NaOH ml total volume ml (c)...
9. In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin...
9. In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin and the excess was then back titrated with HCl. Suppose a 0.4567 g sample of aspirin was reacted with 18.12 ml/of 0.1345 M NaOH and back titrated with 12.15 mL of 0.1124 M HCl to reach the endpoint. Calculate the weight percent of aspirin (Asp) in the sample. 8.457% 37.90% 21.14% 42.28% none of the above
8. Why is it necessary to rid the distilled water of CO,? water oorth Coz or...
8. Why is it necessary to rid the distilled water of CO,? water oorth Coz or dissolve CO2 6 Slightly because dissolued Cla react with H₂O and forms carbonic acid which more water acid So, it will effect the cading of titration experiment.- CO2(g) + H₂ Oce) H₂ CO₂ (aq) carbonic acid. > What is the molarity of a solution that contains 2.38 g of H,, 2H,0 in exactly 300 mL of solution? 10. If 25.21 mL of NaOH solution...
A 0.350-g sample pf an acid, HX, is dissolved in 25.00 mL of water. The resulting...
A 0.350-g sample pf an acid, HX, is dissolved in 25.00 mL of water. The resulting solution is titrated with 0.140 M NaOH solution, and 25.40 mL of the base solution is required to neutralize the acid. Calculate the molar mass of the acid.
A 5.30 mL sample of an H3PO4 solution of unknown concentration is titrated with a 1.100×10-2...
A 5.30 mL sample of an H3PO4 solution of unknown concentration is titrated with a 1.100×10-2 M NaOH solution. A volume of 7.32 mL of the NaOH solution was required to reach the equivalence point. What is the concentration of the unknown H3PO4 solution? Express your answer with the appropriate units.
(1). 500.0 mL of 0.110 M NaOH is added to 535 mL of 0.250 M weak...
(1). 500.0 mL of 0.110 M NaOH is added to 535 mL of 0.250 M weak acid (Ka = 6.08 × 10-5). What is the pH of the resulting buffer? HA(aq) + OH-(aq) = H2O(l) + A- (aq) (2). A 25.863 g sample of aqueous waste leaving a fertilizer )manufacturer contains ammonia. The sample is diluted with 74.612 g of water. A 14.159 g aliquot of this solution is then titrated with 0.1059 M HCl. It required 31.31 mL of...
(1) Boric acid sample (H3BO3) contains boron (Na2B407.10H2O) as impurity. The quantity of this sample was...
(1) Boric acid sample (H3BO3) contains boron (Na2B407.10H2O) as impurity. The quantity of this sample was dissolved in distilled water to give 1 liter of A solution. (a) 25.00 cm3 of solution A required 8.80 cm3 of 0.01 M HCl solution for neutralizing, using methyl red as indicator. (b) In another 25.00 cm3 of A was added 8.80 cm3 of the 0.01 M HCl solution, and some cm3 of glycerol was added. The result is a uniform colorless solution. The...
If a 16.5 mL sample of 1.4 M solution of each of the following acids is reacted with 0.90 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point? (a) 16.5 mL of H,So, titrated with 0.90 M NaOH volume of NaOH mL total volume mL (b) 16.5 mL of HCl titrated with 0.90 M NaOH volume of NaOH ml total volume ml (c)...
9. In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin and the excess was then back titrated with HCl. Suppose a 0.4567 g sample of aspirin was reacted with 18.12 ml/of 0.1345 M NaOH and back titrated with 12.15 mL of 0.1124 M HCl to reach the endpoint. Calculate the weight percent of aspirin (Asp) in the sample. 8.457% 37.90% 21.14% 42.28% none of the above
8. Why is it necessary to rid the distilled water of CO,? water oorth Coz or dissolve CO2 6 Slightly because dissolued Cla react with H₂O and forms carbonic acid which more water acid So, it will effect the cading of titration experiment.- CO2(g) + H₂ Oce) H₂ CO₂ (aq) carbonic acid. > What is the molarity of a solution that contains 2.38 g of H,, 2H,0 in exactly 300 mL of solution? 10. If 25.21 mL of NaOH solution...
(1) Boric acid sample (H3BO3) contains boron (Na2B407.10H2O) as impurity. The quantity of this sample was dissolved in distilled water to give 1 liter of A solution. (a) 25.00 cm3 of solution A required 8.80 cm3 of 0.01 M HCl solution for neutralizing, using methyl red as indicator. (b) In another 25.00 cm3 of A was added 8.80 cm3 of the 0.01 M HCl solution, and some cm3 of glycerol was added. The result is a uniform colorless solution. The...
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