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:
7.0 |
||
> 7.0 |
||
< 7.0 |
||
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
1.1 volt |
||
0.42 volt |
||
1.0 volt |
||
2.2 volt |
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.
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
Calculate the pH of a 25.0 mL of 0.100M base acetic acid solution after being titrated with 0.100 M NaOH to its equivalence point (pKb (acetic acid)=5.68x10^-10)
A solution containing 50.00 mL of 0.100 M of acetic acid is titrated with 0.100 M of a strong base KOH. Calculate the pH of the solution before the base is added. Ka = 18x10-5 OAZO OB 10 OC 29 00.25
A student titrated a 100.0 mL sample of 0.100 M acetic acid with 0.050 M NaOH. (For acetic acid, Ka = 1.8 * 10^-5 at this temperature.) (a) Calculate the initial pH. (b) Calculate the pH after 50.0 mL of NaOH has been added. (c) Determine the volume of added base required to reach the equivalence point. (d) Determine the pH at the equivalence point?
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 sample of 0.100 M acetic acid (Ka = 1.8 × 10−5) in 50.0 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?
40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? [Ka(CH3COOH) = 1.8 × 10–5]
A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4