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Chapter 14: Problem 171

Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7}-M\) solution of \(\mathrm{NaOH}\) in water.

Short Answer

Expert verified
The pH of the \(1.0 \times 10^{-7}-M\) NaOH solution is 7.

Step by step solution

01

Determine the concentration of hydroxide ions (OH⁻)

Since NaOH is a strong base, it will dissociate completely in water. The equation for this dissociation is: NaOH → Na⁺ + OH⁻ Given the concentration of NaOH is \(1.0 \times 10^{-7} \, M\). As it dissociates completely in water, the concentration of OH⁻ ions will be equal to the concentration of NaOH. Thus, [OH⁻] = \(1.0 \times 10^{-7} \, M\)
02

Calculate the pOH of the solution

The pOH is calculated as the negative base-10 logarithm of the hydroxide ion concentration: pOH = -log10([OH⁻]) Here, [OH⁻] = \(1.0 \times 10^{-7} \, M\). So, we can plug this value into the formula: pOH = -log10(\(1.0 \times 10^{-7}\)) = 7
03

Calculate the pH of the solution

We know the relationship between pH and pOH is given by: pH + pOH = 14 Since we have already calculated the pOH as 7, we can find the pH as follows: pH = 14 - pOH = 14 - 7 = 7 Therefore, the pH of the \(1.0 \times 10^{-7}-M\) NaOH solution is 7.

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Most popular questions from this chapter

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100-M\) solution.

Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation \(2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\).

Rank the following \(0.10 M\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\)
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