From the calculations, the pH of the final solution is 9.04.
<h3>What is the pH of the buffer?</h3>
We can use the Henderson Hasselbach equation to obtain the final pH of the solution in terms of the pKb and the base concentration.
Number of moles of salt = 250/1000 L * 0.5 M = 0.125 moles
Number of moles of base = 150/1000 L * 0.5 M = 0.075 moles
Total volume of solution = 250ml + 150ml = 400ml or 0.4 L
Molarity of base = 0.075 moles/ 0.4 L = 0.1875 M
Molarity of salt = 0.125 moles/ 0.4 L = 0.3125 M
pOH = pKb + log[salt/base]
pKb = -log(1.8 x 10^-5) = 4.74
pOH = 4.74 + log[0.3125/0.1875 ]
pOH = 4.96
pH = 14- 4.96
pH = 9.04
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because there are 12 atoms of hydrogen in a molecule of glucose, multiply 12 by Avogadro's number (6.02x10^23) to get how many molecules of hydrogen there are in a mole of glucose. Then multiply that number by 1.75, which is the number of moles of glucose there is in this problem.
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Explanation:
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H2 + O2
Explanation:
H2 + O2 --> H2O
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