Question

In: Chemistry

what is the concentration of H+ ions after adding 5.41 mL of of 0.816 M NaOH...

what is the concentration of H+ ions after adding 5.41 mL of of 0.816 M NaOH to 65.7 mL of 1.97 M HNO3?

Solutions

Expert Solution

NaOH                  + HNO3 -----> NaNO3(aq)    + H2O

5.41x0.816                65.7x1.97                    0                         0     initial concentrations

=4.414                      =129.429

0                                125.015                       4.414              4.414    after reaction

Thus [H+] = 125.015/71.11= 1.758 M

The H+ ion concentration in the solution is 1.758M


Related Solutions

What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added...
What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added to 120 mL of 0.200 M Na2SO4? What is the concentration of hydroxide ions after 50.0 mL of 0.250 M NaOH is added to 120 mL of 0.200 M Sr(OH)2?
What is the pOH of a solution obtained by adding 36.7 mL of 0.300 M NaOH...
What is the pOH of a solution obtained by adding 36.7 mL of 0.300 M NaOH (aq) to 44.3 mL of 0.200 M CH3COOH (aq)? Select one: a. 2.96 b. 2.33 c. 1.58 d. 1.85 QUESTION What is the total volume of solution at the equivalence point if a 0.100 M NaOH (aq) solution is used to titrate 15.0 mL of 0.300 M HClO4 (aq)? Select one: a. 75.0 mL b. 55.5 mL c. 45.0 mL d. 60.0 mL
What is the pH of a solution obtained by adding 50.0 mL of 0.100 M NaOH...
What is the pH of a solution obtained by adding 50.0 mL of 0.100 M NaOH (aq) to 60.0 mL of 0.100 M HCl (aq)? Select one: a. 11.96 b. 2.04 c. 3.11 d. 7.00 QUESTION 2 What is the pH of a solution prepared by dissolving 5.86 grams of propanoic acid, CH3CH2COOH (l), and 1.37 grams of NaOH (s) in enough water to make exactly 250.0 mL of solution. pKa = 4.87 for propanoic acid? Select one: a. 4.58...
A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of...
A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH2PO4. How many moles of HCl must be added to this buffer solution to change the pH by 0.25 units? Assume the total volume remains unchanged at 400 mL. For H2PO4-, Ka = 6.3 × 10-8.
What is the pH after adding 10.0 mL of 0.10 M HCl to a 90.0 mL...
What is the pH after adding 10.0 mL of 0.10 M HCl to a 90.0 mL buffer solution of 0.60 M CH3COOH and 0.40 M CH3COONa at 25 °C? (Ka = 1.8 x 10−5)? What reaction will occur when HCl is added to the above buffer? 1) How many moles of acetic acid are present after the reaction? 2) How many moles of acetate ion are present after the reaction? 3) What is the concentration (in M) of acetic acid...
You prepare a diluted NaOH solution by adding 15 mL of 3 M NaOH to 435...
You prepare a diluted NaOH solution by adding 15 mL of 3 M NaOH to 435 mL of distilled water and stir for several minutes. In a clean dry beaker you obtain about 75 mL of standardized HCl and record the molarity. You take two burets, rinse one with the standardized HCl and one with the diluted NaOH solution. Fill one buret with the HCl and the other with NaOH. Prior to the titration the standard HCl soltion (~15.00 mL)...
What is the pH after the addition of 20.0 mL of 0.10 M NaOH to 80.0...
What is the pH after the addition of 20.0 mL of 0.10 M NaOH to 80.0 mL of a buffer solution that is 0.25 M in NH3 and 0.25 M in NH4Cl, (K b =1.8 x 10 −5 for NH3 )?
at 25 C, a solution is prepared by adding 50 ml of 0.2 M NaOH to...
at 25 C, a solution is prepared by adding 50 ml of 0.2 M NaOH to 75 ml of 0.1 M NaOH. What is the total concentratoin of hydorxide ion in the solution? Assume volumes are additive.
What is the initial pH, at half way and after 12.5 mL of NaOH 0.3235 M...
What is the initial pH, at half way and after 12.5 mL of NaOH 0.3235 M is used to titrate 65.00 mL of 0.1237 M HAc?    Ka=1.75 x 10-5
determine the pH at each stage: (a) before adding 0.50 M of NaOH to 25.0 mL...
determine the pH at each stage: (a) before adding 0.50 M of NaOH to 25.0 mL of 0.45 M HCl (b) after adding 15.00 mL of NaOH to 25.00 mL of 0.45 M of HCl. (c) at the equivalent point (D) after adding 35.00 mL 0.50 M of NaOH to 25.00 mL of 0.45 M HCl
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT