Question

1)Design a buffer that has a pH of **10.12** using
one of the weak base/conjugate acid systems shown below.

Weak Base | K_{b} |
Conjugate Acid | K_{a} |
pK_{a} |
---|---|---|---|---|

CH |
4.2×10 |
CH |
2.4×10 |
10.62 |

C |
5.9×10 |
C |
1.7×10 |
7.77 |

C |
1.5×10 |
C |
6.7×10 |
5.17 |

How many grams of the **bromide** salt of the
conjugate acid must be combined with how many grams of the weak
base, to produce **1.00** L of a buffer that is
**1.00** M in the weak base?

grams **bromide** salt of conjugate acid =

grams weak base =

2)Design a buffer that has a pH of **10.91** using
one of the weak acid/conjugate base systems shown below.

Weak Acid | Conjugate Base | K_{a} |
pK_{a} |
---|---|---|---|

HC |
C |
6.4×10 |
4.19 |

H |
HPO |
6.2×10 |
7.21 |

HCO |
CO |
4.8×10 |
10.32 |

How many grams of the **potassium** salt of the weak
acid must be combined with how many grams of the
**potassium** salt of its conjugate base, to produce
**1.00** L of a buffer that is **1.00**M
in the weak base?

grams **potassium** salt of weak acid
= g

grams **potassium** salt of conjugate base
= g

Will rate answer!

Answer #1

According to Henderson-Hasselbulch equation:

pH = pKa + Log([base]/[acid])

i.e. 10.12 = 10.62 + Log([CH3NH2]/[CH3NH3^{+}])

i.e. Log([CH3NH3^{+}]/[CH3NH2]) = 10.62 - 10.12 =
0.5

i.e. [CH3NH3^{+}]/[CH3NH2] = 10^{0.5} =
3.162

i.e. **The percent of base (CH3NH2) in the buffer**
= 100/{1 + ([CH3NH3^{+}]/[CH3NH2])}

= 100/(1+3.162) ~ **24%**

Therefore, the **percent of acid (CH3NH3 ^{+}) in
the buffer** = 100 - 24 =

1 M - 24%

i.e. 76% = (76/24) M = 3.167 M

Therefore, the mass of weak base (CH3NH2) = 1 mol * 31 g/mol =
**31 g**

And the mass of bromide salt of conjugate acid (CH3NH3Br) =
3.167 mol * 112 g/mol = **354.7 g**

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pKa
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4.2×10-4
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10.62
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5.9×10-7
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1.7×10-8
7.77
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1.5×10-9
C5H5NH+
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