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Chapter 17: Problem 38

Two solutions are mixed: \(100.0 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) with \(\mathrm{pH} 2.50\) and \(100.0 \mathrm{mL}\) of \(\mathrm{NaOH}(\text { aq) with } \mathrm{pH} 11.00\) What is the pH of the resulting solution?

Short Answer

Expert verified
The pH of the resulting solution can be calculated by first determining the concentrations of the \(H^{+}\) ions and \(OH^{-}\) ions in the mixed solution, and then using this information to calculate the resulting pH. The specific values are subject to the calculated numerical values.

Step by step solution

01

Calculate the concentration of Hydronium ions

First, use the provided pH of 2.50 for Hydrochloric acid (HCl) to find the Hydronium ion concentration. This can be done using the formula \(H^{+} = 10^{-\text{pH}}\). Hence the \(H^{+}\) concentration will be \(10^{-2.5}\).
02

Calculate the concentration of Hydroxide ions

Next, find the Hydroxide ion concentration using the pH of 11 for Sodium Hydroxide (NaOH). However firstly we need to convert this to pOH using formula \(\text{pOH} = 14 - \text{pH}\) which gives us the pOH as 3.00. Then, find the \(OH^{-}\) ion concentration, similar as before, with formula \(OH^{-} = 10^{-pOH}\), giving \(10^{-3}\).
03

Find the concentrations after dilution

The solution has been diluted by mixing 100ml of HCl with 100ml of NaOH, to give a 200ml solution. The acid and base ion concentrations will thus be halved, resulting to \(H^{+} = 10^{-2.5}/2\) and \(OH^{-} = 10^{-3}/2\).
04

Establish and solve the neutralization reaction

In solution, there will be a neutralization reaction, where \(H^{+}\) from HCl reacts with \(OH^{-}\) from NaOH to form water. As per the concentrations we have, it is evident that there is excess \(H^{+}\). Therefore, we have to compute the remaining \(H^{+}\) concentration after all the \(OH^{-}\) ions have reacted and formed water. This will be \((10^{-2.5}/2) - (10^{-3}/2)\).
05

Calculate the pH of the resulting solution

The final task is finding the pH of the resulting solution, which can be achieved by using \(\text{pH} = -\log[\(H^{+}\)]\) on the remaining \(H^{+}\) concentration.

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Most popular questions from this chapter

Sodium ammonium hydrogen phosphate, \(\mathrm{NaNH}_{4}\) \(\mathrm{HPO}_{4},\) is a salt in which one of the ionizable \(\mathrm{H}\) atoms of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is replaced by \(\mathrm{Na}^{+},\) another is replaced by \(\mathrm{NH}_{4}^{+},\) and the third remains in the anion \(\mathrm{HPO}_{4}^{2-}\) Calculate the \(\mathrm{pH}\) of \(0.100 \mathrm{M} \mathrm{NaNH}_{4} \mathrm{HPO}_{4}(\mathrm{aq})\) [Hint: You can use the general method introduced on page \(720 .\) First, identify all the species that could be present and the equilibria involving these species. Then identify the two equilibrium expressions that will predominate and eliminate all the species whose concentrations are likely to be negligible. At that point, only a few algebraic manipulations are required.]

Sulfuric acid is a diprotic acid, strong in the first ionization step and weak in the second \(\left(K_{\mathrm{a}_{2}}=1.1 \times 10^{-2}\right)\) By using appropriate calculations, determine whether it is feasible to titrate \(10.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\)

Phenol red indicator changes from yellow to red in the pH range from 6.6 to \(8.0 .\) Without making detailed calculations, state what color the indicator will assume in each of the following solutions: (a) \(0.10 \mathrm{M} \mathrm{KOH}\) (b) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (c) \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3} ;\) (d) \(0.10 \mathrm{M}\) HBr; (e) \(0.10 \mathrm{M} \mathrm{NaCN} ;\) (f) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO}\).

To convert \(\mathrm{NH}_{4}^{+}(\text {aq })\) to \(\mathrm{NH}_{3}(\mathrm{aq}),\) (a) add \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) raise the \(\mathrm{pH} ;\) (c) add \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (d) add \(\mathrm{NaCl}\).

The \(\mathrm{pH}\) of ocean water depends on the amount of atmospheric carbon dioxide. The dissolution of carbon dioxide in ocean water can be approximated by the following chemical reactions (Henry's Law constant for \(\left.\mathrm{CO}_{2} \text { is } K_{\mathrm{H}}=\left[\mathrm{CO}_{2}(\mathrm{aq})\right] /\left[\mathrm{CO}_{2}(\mathrm{g})\right]=0.8317 .\right)\) \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq})\) \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{-}(\mathrm{aq})\) \(\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{HCO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\) (a) Use the equations above to determine the hydronium ion concentration as a function of \(\left[\mathrm{CO}_{2}(\mathrm{g})\right]\) and \(\left[\mathrm{Ca}^{2+}\right]\) (b) During preindustrial conditions, we will assume that the equilibrium concentration of \(\left[\mathrm{CO}_{2}(\mathrm{g})\right]=280\) ppm and \(\left[\mathrm{Ca}^{2+}\right]=10.24 \mathrm{mM} .\) Calculate the \(\mathrm{pH}\) of a sample of ocean water.
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