Write the net ionic equation for the CaH_2/H_2O reaction in Part B. Write the ne
ID: 507888 • Letter: W
Question
Write the net ionic equation for the CaH_2/H_2O reaction in Part B. Write the net ionic equation for the products of the reaction in question 1 reacting with the solution you chose in Part B (NaC_2H_3O_2, NaCl, NaHSO_4). A few mL of ___ M HCl are added to 20 mL of ____ M NaC_2H_3O_2: a. Will the initial pH of the _____ M NaC_2H_3O_2 be greater than, or equal to 7? A. Greater than 7 B. Less than 7 C. Equal to 7 b. As HCl is added to the ___ M NaC_2H_3O_2, will the pH of the solution increase, decrease, or remain unchanged? A. Increase B. Decrease C. Remain unchanged c. write the net ionic equation for the reaction between the HCl solution and the NaC_2H_3O_2 solution A solution has a hydroxide concentration of _____ M. Calculate: a. pH b. pOH c. [H^+]Explanation / Answer
Answer:
1. The net ionic reaction of CaH2/H2O is
CaH2(s) + 2 H2O(l) Ca2+(aq) + 2 OH-(aq) + 2 H2(g).
OR
CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
2.Ca(OH)2 reacts with NaC2H3O2, NaCl and NaHCO3
a.Ca(OH)2 + 2NaC2H3O2------->Ca(C2H3O2)2 + 2Na(NO3)
b.Ca(OH)2 + 2 NaCl ---------> CaCl2 + 2 NaOH
c.NaHCO3 + Ca(OH)2 ------>CaCO3 + H2O + NaOH
NaHCO3 + Ca(OH)2 --------->CaCO3 + H2O + Na2CO3
3.
a. Initia pH of 0.15 M NaC2H3O4 is less than 7
C2H3O2- in water --> HC2H3O2 & OH-
K = [HC2H3O2] [OH-] / [C2H3O2-]
K hydrolysis = K water / K HC2H3O2 = 1 X 1014 / 1.8 X 10-5 = 5.55 X 10-10
5.55 X 10-10 = [x] [x] / [0.15]
x = [OH-] = 8.23 X 10-11 molar
pOH = 10.07
pH = 14 - pOH
pH = 3.92( Acidic)
b. HCl is added to NaC2H3O2 pH of solution decreases because following acid formation reaction occur
HCl + NAC2H3O2 ---------> HC2H3O2 + NaCl
c.Reaction of HCl with NaC2H3O2
HCl + NAC2H3O2 ---------> HC2H3O2 + NaCl
4 . [OH-] =0.010
a. pOH = -log[OH-]
pOH=-log[0.010]
=2
pH + pOH =14
pH=12
b.
pOH = -log[OH-]
pOH=-log[0.010]
=2
c.
pH=12
pH=-log[H+]
[H+]=1 X 10-12
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