Formulas: pH = -log ([H3O+]) [H3O+] = 10-pH Kw = [H3O+][OH-] = 1.0 x 10-14 1. Wr

Question

Formulas:

pH = -log ([H3O+])

[H3O+] = 10-pH

Kw = [H3O+][OH-] = 1.0 x 10-14

1. Write the conjugate bases for these substances which are acids:

HSe- HOBr PH4+   H2ASO4-

1b. Write the conjugate acids for these substances which are bases:

TeO32-   AsH3 H2PO4-   BrO2-

2. Indicate whether the following solutions are acidic, basic or neutral: {Note: you can predict the answers without doing any calculations}

(a) [H3O+] = 7.20 x 10–5 M(d) [OH-] = 1.0 x 10–7 M

(b) [H3O+] = 5.30 x 10–10 M        (e) pH = 2.22

(c) [OH-] = 3.50 x 10–3 M (f) pH = 8.78

Explanation / Answer

1) HSe-  is acid conjugate base = Se-2

HOBr is acid conjugate base = OBr-

PH4+  is acid conjugate base = PH3

H2AsO4-  is acid conjugate base HAsO4-2

1 b) TeO3-2  base conjugate acid = HTeO3-2

AsH3 is base conjugate acid = AsH4+  

H2PO4-  is base conjugate acid = H3PO4

BrO2-  is base conjugate acid = HBrO2

2) [H3O+] = 7.20 x 10-5 M

pH = - log [7.20 x 10-5]

pH = 4.14

solution is acidic

b) [H3O+] = 5.30 x 10-10 M

pH = - log [5.30 x 10-10]

pH = 9.27

solution is basic.

c) [OH-] = 3.50 x 10-3 M

pOH = - log [3.50x 10-3]

pOH = 2.46

pH = 14 - 2.46

pH = 11.54

solution is basic in nature.

d) [OH-] = 1.0 x 10-7 M

pOH = - log [1.0 x 10-7]

pOH = 7.0

pH = 14 - 7.0

pH = 7.0

solution is neutral.

e) pH = 2.2 solution is acidic

f) pH = 8.78

solution is basic

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