If any of these questions can be answered, I\'d appreciate it. 1. Calculate the

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If any of these questions can be answered, I'd appreciate it.

1. Calculate the pH of each of the following solutions: (use tables in text....p. 391)
a) 0.60 M HCl b) 0.40 M KOH c) 0.75 M HNO2 d) 0.88 M NaF e) 15 M HNO3 f) 0.95 M NH4Cl

2. Calculate the pH of the resulting solutions: a) 35.0 ml of 0.300 M HCl added to 20.0 ml 0.500M NaOH b) 30.0 ml of 0.545 M KOH added to 30.0 ml of 0.545 M HC2H3O2
c)55.0 ml of 0.880 M HCl added to 40.0 ml of 1.21 M NaC2H3O2

3. Calculate the percent dissociation in #1 (f) above.

4. Calculate the resulting pH when you titrate 33.0 ml of HCl to the equivalence point with 0.580 M NH3. ( [HCl] = 0.750 M )

5. Which indicator would you use to determine the equivalence point if you performed a titration using the solutions in #2 & #4 ?

6. In the standardization of 22.0 ml of NaOH solution, it was determined that 14.5 ml of 0.335 M HCl was required to reach the equivalence point. What is the molarity of the NaOH solution? What is the pH at this point?

7. How many grams of NaOH are required to completely neutralize 32.0 ml of 0.678 M HNO3?

8. What is the [OH-] in #1 (a) above?

9. What is the pOH in #1 (d) above?

10. Label the following as acidic, basic (alkaline), or neutral: (all aqueous solutions) a) NaNO3 b) NaCN c) ZnCl2 d) Al(NO3)3 e) K2C3O4
f) NH3 g) NH4CN

Explanation / Answer

1) for strong acids u can directly find pH as -log(c) 2) part a). HCl + NaOH -----> NaCl + H2O If equal amounts and molarities of NaOH and HCl were added they would make water which is neutal. This is not the case so we have to find out how much of one will be left over. C (HCl) = n/v 0.300 = n(HCl) /0.035 n(HCl) = 0.0105 C (NaOH) = n/v 0.500 = n/0.02 n(NaOH) = 0.01 mol As you can see there is more HCl than NaOH. n(HCl) left over = 0.0105-0.01 = 0.0005 mol We can now calculate the pH n(HCl) = c/v 0.0005 = c/0.055

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