student titrated 15.0 mL of acetic acid acid, HNO3 solution with 25.0 mL of a 0.100 M NH4OH solution.The PH at equivalent point is:
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7.0 |
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> 7.0 |
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< 7.0 |
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8.5 |
2.
The standard cell potential for the voltaic cell based on the reaction below is __________ V.
Cu 2+ + 2 e ------> Cu 0 E,Cu= -0.34 Volt
Zn 2+ + 2 e ------> Zn 0E, Zn = 0.76 Volt
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1.1 volt |
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0.42 volt |
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1.0 volt |
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2.2 volt |
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student titrated 15.0 mL of acetic acid acid, HNO3 solution with 25.0 mL of a 0.100 M...
V. The standard cell potential for the voltaic cell based on the reaction below is Cu 2+ + 2 e -----> Cu °E,Cu= -0.34 Volt + 2 e------> Zn OE, Zn = 0.76 Volt Zn2+ 1.1 volt 0.42 volt 1.0 volt 2.2 volt
A student titrated 15.0 mL of acetic acid acid, HC2H302 ( Molar mass= 60 g/mol) solution with 25.0 mL of a 0.100 M NaOH solution The Molarity(mol/L) of HC2H302 required for complete titration is? HC2H302 + NaOH - NaC2H302 + H20 A) 0.0250 M B) 0.250 M C) 2.5M D) 0.0167M E) 0.167 mol
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A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.1007 M NaOH and the equivalence point volume was determined by graphical means to be 25.42 mL. What is the concentration of the acetic acid? concentration of HA = ___ M
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A sample of 0.100 M acetic acid (K, - 18*10*5) in 300 ml. of solution is titrated with standard 0.100 M NaOH. What is the pH in the titration flask after addition of 15.0 ml. of NaOH? 02.18 On30 Oct. oo Od 11.40 De 47
A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10-5? Final Answer: _______
a. A 25.0 mL of a solution of acetic acid (HA) of unknown concentration is titrated with 0.0919 M NaOH and the equivalence point volume was determined by graphical means to be 25.91 mL. What is the concentration of the acetic acid? concentration of HA = ___ M b. Calculate the pKa of a weak monoprotic acid if the pH of a 0.162 M solution is 2.52. pKa = ___