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Answer is: pH of a buffer is 4.64.
ck(CH₃COOH) = 0.45 M.
cs(CH₃COONa) = 0.35 M.
Ka = 1.8·10⁻⁵.
<span>pKa = -logKa.
</span>pKa = -log(1.8·10⁻⁵) = 4.75.
<span>Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
</span>pH = 4.75 + log(0.35M/0.45M).
pH = 4.75 - 0.11.
pH = 4.64.
pH (potential of hydrogen) is a numeric
scale used to specify the acidity or basicity an aqueous solution.
Answer:
The products are Al(NO₃)₃(aq) and H₂(g).
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)
Explanation:
Let's consider the reaction between aluminum and nitric acid. This is a single replacement reaction, in which Al replaces H in HNO₃ to form aluminum nitrate. The unbalanced reaction is:
Al(s) + HNO₃(aq) ⇒ Al(NO₃)₃(aq) + H₂(g)
We start balancing N atoms by multiplying HNO₃ by 3.
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + H₂(g)
Finally, we get the balanced equation multiplying H₂ by 1.5.
Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)
MH₂: 2 g/mol
mNH₃: 17 g/mol
...............................
3H₂ + N₂ ---> 2NH₃
6g......................34g
6g H₂ ---- 34g NH₃
Xg -------- 13,09g
X = (6×13,09)/34
X = 2,31g H₂