Answer:
pH change is -0.07
Explanation:
Using H-H equation for acetic acid:
pH = pKa + log [Acetate salt] / [Acetic acid]
Replacing:
pH = 4.74 + log[1.188M] / [1.188M]
pH = 4.74
The HCl reacts with sodium acetate producing acetic acid, thus:
HCl + CH₃COONa → CH₃COOH + NaCl
That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.
As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:
sodium acetate: 1.188mol - 0.1mol = 1.088mol
Acetic acid: 1.188mol + 0.1mol = 1.288mol
Using again H-H equation:
pH = 4.74 + log[1.088M] / [1.288M]
pH = 4.67
pH change is: 4.67 - 4.74 = -0.07
17.8 mL NaOH
<em>Step 1.</em> Write the chemical equation
Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)
<em>Step 2.</em> Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]
= 11.50 mmol Fe^(2+)
<em>Step 3.</em> Calculate the moles of NaOH
Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]
= 23.00 mmol NaOH
<em>Step 4.</em> Calculate the volume of NaOH
Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)
= 17.8 mL NaOH
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Answer:
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