100.0 mL of 3.00 M HCl (in water) is added to 50.0 mL of 5.00 M NaOH in a beaker. What is
present in the beaker?
HCl + NaOH --------------> NaCl + H2O
millimoles of NaOH = 50.0 x 5.00 = 250
millimoles of HCladded = 100.0 x 3.0 = 300
300 - 25 = 50.0 millimoles NaOH left in the solution
total volume = 100 + 50 = 150 mL
[NaOH] = 50 / 150 = 0.33 M
resulting NaOH = 0.33 M
As 50.0 mL of 0.10 M HCl is added to 100.0 mL of 0.10 M NaOH, what happens to the pH of the NaOH solution?
Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). NH3 Kb=1.8x10^-5 Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH, (aq) and 0.100 M in NHCl(aq). Consult the table of ionization constants as needed. ApH = Calculate the change in pH when 3.00 mL of 0.100 M NaOH is added to the original buffer solution. ApH =
As 50.0 mL of 0.10 M NaOH is added to 100.0 mL of 0.5 M NaOH, what happens to the pH of the original solution? does it increase or decrease
3. (a) Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). (b) Calculate the change in pH when 5.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3 (aq) and 0.100 M in NH4CI(aq). Consult the table of ionization constants as needed ДрН Calculate the change in pH when 3.00 mL of 0.100 M N2OH is added to the original buffer solution. АрН -
When 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH are mixed in a constant-pressure calorimeter, the temperature of the solution increases from 21.0°C to 27.5°C. Calculate the enthalpy change of the reaction per mole of HCl assuming the solution has a total volume of 100.0 mL and a density of 1.000 g/mL. The specific heat of water is 4.184 J/q°C Asoln = 2720
If you mix 100.0 mL of 0.125 M HCl (strong acid) with 50.0 mL of 0.175 M NaOH (strong base) what will be the pH of the resulting solution? Have you reached the endpoint of the reaction (circle your answer)? YES NO Explain your answer:
When 50.0 mL of .10 M HCl and 50.0 mL of .10 M NaOH, both at 22 oC, are added to a calorimeter, the temperature of the mixture reaches 28.9 oC. Calculate the heat produced by this reaction. Density of water 1.00g/mL. Specific heat of water = 4.184 J/g oC
If 5.0 mL of 0.1 M NaOH is added to 50.0 mL of 0.1 M HCl, what will be the resulting pH of the solution? Round your answer to the tenths place. Do not include units in your response.