Enter the absolute value of the change in pH caused by the addition of 1.0 mL of 0.10 M NaOH to each of the solutions. pH of H2O + NaOH: 11.59 pH WA + NaOH: 3.96 pH CB + NaOH: 11.44 pH Buffer + NaOH: 4.90 pH
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The initial pH of pure water (H2O) is 7. When 1.0 mL of 0.10 M NaOH is added, the pH becomes 11.59. So, the change in pH for H2O + NaOH is |11.59 - 7| = 4.59. The pH of weak acid (WA) before the addition of NaOH is not given, so we cannot calculate the change Show more…
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Calculate the $\mathrm{pH}$ change when $10.0 \mathrm{~mL}$ of $0.10-\mathrm{M} \mathrm{NaOH}$ is added to $90.0 \mathrm{~mL}$ pure water, and compare the $\mathrm{pH}$ change with that when the same amount of NaOH solution is added to $90.0 \mathrm{~mL}$ of a buffer consisting of $1.0-\mathrm{M} \mathrm{NH}_{3}$ and 1.0-M $\mathrm{NH}_{4} \mathrm{Cl}$. Assume that the volumes are additive. $K_{\mathrm{b}}$ of $\mathrm{NH}_{2}=1.7 \times 10^{-5}$
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