Please also explain how you calculate the initial moles, because I do not get the correct number.Calculate the $\mathrm{pH}$ of the resulting solution at 25 degrees $\mathrm{C}$, when $50,1 \mathrm{ml}$ of $0,100 \mathrm{~mol} /$ sodium hydroxide solution $\mathrm{NaOH}$ has been added to $50,0 \mathrm{ml}$ of;
a) $0,100 \mathrm{~mol} / \mathrm{I}$ hydrochloric acid $\mathrm{HCl}$ solution.
b) $0,100 \mathrm{~mol} / \mathrm{l}$ ethanoic acid $\mathrm{CH} 3 \mathrm{COOH}$ solution. $\mathrm{pKa}$ of ethanoic acid $=4,76$.

Question

Please also explain how you calculate the initial moles, because I do not get the correct number.Calculate the $\mathrm{pH}$ of the resulting solution at 25 degrees $\mathrm{C}$, when $50,1 \mathrm{ml}$ of $0,100 \mathrm{~mol} /$ sodium hydroxide solution $\mathrm{NaOH}$ has been added to $50,0 \mathrm{ml}$ of;
a) $0,100 \mathrm{~mol} / \mathrm{I}$ hydrochloric acid $\mathrm{HCl}$ solution.
b) $0,100 \mathrm{~mol} / \mathrm{l}$ ethanoic acid $\mathrm{CH} 3 \mathrm{COOH}$ solution. $\mathrm{pKa}$ of ethanoic acid $=4,76$.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate initial moles of NaOH**: To find the initial moles of NaOH, use the concentration and volume of the NaOH solution. The formula is $n = C imes V$, where $n$ is the number of moles, $C$ is the concentration in mol/L, and $V$ is the volume in liters. ‖ ‖ 2 ⋮ **Calculate initial moles of HCl**: Similarly, calculate the initial moles of HCl using its concentration and volume. ‖ ‖ 3 ⋮ **Reaction between NaOH and HCl**: NaOH and HCl react in a 1:1 molar ratio to form water and sodium chloride. The reaction is $ \mathrm{NaOH} + \mathrm{HCl} ightarrow \mathrm{NaCl} + \mathrm{H_2O}$. ‖ ‖ 4 ⋮ **Determine the limiting reactant**: Since both NaOH and HCl have the same molarity and nearly the same volume, they will completely neutralize each other. ‖ ‖ 5 ⋮ **Calculate the pH**: After neutralization, the pH of the resulting solution will be neutral, which is 7 at 25°C. ‖ ∻Answer∻ ⚹ The pH of the resulting solution after adding NaOH to HCl is 7. ⚹ ∻Key Concept∻ ⚹ Neutralization reaction between a strong acid and a strong base results in a neutral pH. ⚹ ∻Explanation∻ ⚹ Since NaOH and HCl react in a 1:1 ratio and their initial moles are equal, they completely neutralize each other, resulting in a neutral pH of 7. ⚹ --- ∻Solution∻ ‖ 1 ⋮ **Calculate initial moles of NaOH**: Use the concentration and volume of the NaOH solution to find the initial moles. ‖ ‖ 2 ⋮ **Calculate initial moles of CH3COOH**: Calculate the initial moles of ethanoic acid using its concentration and volume. ‖ ‖ 3 ⋮ **Reaction between NaOH and CH3COOH**: NaOH and CH3COOH react in a 1:1 molar ratio. The reaction is $ \mathrm{NaOH} + \mathrm{CH_3COOH} ightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$. ‖ ‖ 4 ⋮ **Determine the limiting reactant**: Since both NaOH and CH3COOH have the same molarity and nearly the same volume, they will completely neutralize each other. ‖ ‖ 5 ⋮ **Calculate the pH of the buffer solution**: After neutralization, the solution will contain the salt CH3COONa, which hydrolyzes to form CH3COO- and Na+. The CH3COO- will react with water to form CH3COOH and OH-. Use the Henderson-Hasselbalch equation to calculate the pH: $ \mathrm{pH} = \mathrm{pKa} + \log \left( \frac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} ight) $. Since the moles of CH3COOH and CH3COO- are equal, the ratio is 1, and the log of 1 is 0. ‖ ∻Answer∻ ⚹ The pH of the resulting solution after adding NaOH to CH3COOH is 4.76. ⚹ ∻Key Concept∻ ⚹ The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation. ⚹ ∻Explanation∻ ⚹ After neutralization, the solution acts as a buffer with equal concentrations of CH3COOH and CH3COO-, resulting in a pH equal to the pKa of ethanoic acid. ⚹◊How can the initial moles of sodium hydroxide and the other acid solution be calculated in this scenario?⍭ Generate me a similar question◊

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