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Allyson F.

Former Teacher's Assistant for Various STEM Courses

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Chemistry

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Question:

Determine the pH of a solution that is 0.0166 M in $$Na_{2}SO_{4}$$, making sure to consider the activities of spectator ions and the ionic strength of the solution. Use the following values to find the activity coefficients for some selected ions: If the solution has an ionic strength of 0.010, 0.050, or 0.100, then hydrogen ion, $$H^{+}$$, will have an activity coefficient of 0.914, 0.860, or 0.830, respectively, and hydroxide ion, $$OH^{-}$$, will have an activity coefficient of 0.900, 0.810, or 0.760, respectively.

Allyson F.

Answer:

When sodium sulfate, $$Na_{2}SO_{4}$$, is dissolved in water, it dissociates into its respective ions of $$Na^{+}$$ and $$SO_{4}^{2-}$$. We must first find the concentrations of these ions in the solution. For the concentration of sodium cation, $$\frac{0.0166\;mol\:Na_{2}SO_{4}}{1\:L}\;\times\;\frac{2\;mol\;Na^{+}}{1\;mol\;Na_{2}SO_{4}}=0.0332\;M\;Na^{+}$$ For the concentration of sulfate anion, $$\frac{0.0166\; mol\:Na_{2}SO_{4}}{1\:L}\;\times\;\frac{1\;mol\;SO_{4}^{2-}}{1\;mol\;Na_{2}SO_{4}}=0.0166\;M\;SO_{4}^{2-}$$ Now that we have calculated the concentrations of each ion in the solution, we can now find the ionic strength of the solution: $$I=\frac{1}{2}\sum_{i=1}^{N}c_iz_i^2=\frac{1}{2}[(0.0332\;M\;Na^{+})(+1)^2+(0.0166\;M\;SO_4^{2-})(-2)^2]=0.0498\;M\approx0.05\;M$$ Since the ionic strength of the solution is approximately 0.05 M, we will refer to that column in the table for activity coefficients. Thus, $$\gamma_{H_3O^+}=0.860$$ $$\gamma_{OH^-}=0.810$$ In order to see how the "spectator" ions of sodium and sulfate affect the pH of the solution, we must consider the ionization (or dissociation) of water: $$2H_{2}O\;(l)\rightleftharpoons H_{3}O^{+}\;(aq)+OH^{-}\;(aq)$$. If we set a corresponding ICE chart with this reaction, beginning with 0 M for the concentrations of the products, we will end up with $$x$$ molar concentrations of both hydronium and hydroxide ions. Since pH is determined by the concentration of hydronium ions present in the solution, we must find that concentration, and we can do so by using the dissociation constant definition for water: $$K_w=\alpha_{H_3O^+}\cdot \alpha_{OH^-}=\gamma_{H_3O^+}\cdot \gamma_{OH^-}[H_3O^+][OH^-]=(0.860)x(0.810)x=0.6966x^2=1.0\times 10^{14}$$ $$x=[H_3O+]=1.2\times10^{-7}\;M$$ Now that we have the concentration for water, we must find its activity denoted by $$\alpha_{H_3O^+}$$: $$\alpha=\gamma\cdot[H_3O^+]=(0.860)(1.2\times10^{-7})=1.030\times10^{-7}$$ Now, we can finally calculate the pH of the sodium sulfate solution: $$pH=-log\alpha_{H_3O^+}=-log(1.03\times10^{-7})\approx6.987$$

Chemistry

TutorMe

Question:

Determine the pH of a solution that is 0.0166 M in $$Na_{2}SO_{4}$$, making sure to consider the activities of spectator ions and the ionic strength of the solution. Use the following table to find the activity coefficients for some selected ions: $$\begin{tabular}{ |p{3cm}||p{3cm}|p{3cm}|p{3cm}| } \hline \multicolumn{4}{|c|}{Ionic Strength} \\ \hline Ion & 0.010 & 0.050 & 0.100 \\ \hline H+ & 0.914 & 0.860 & 0.830 \\ OH- & 0.900 & 0.810 & 0.760 \\ \hline \end{tabular}$$

Allyson F.

Answer:

When sodium sulfate, $$Na_{2}SO_{4}$$, is dissolved in water, it dissociates into its respective ions of $$Na^{+}$$ and $$SO_{4}^{2-}$$. We must first find the concentrations of these ions in the solution. For the concentration of sodium cation, $$\frac{0.0166\; mol\:Na_{2}SO_{4}}{1\:L}\;\times\;\frac{2\;mol\;Na^{+}}{1\;mol\;Na_{2}SO_{4}}=0.0332\;M\;Na^{+}$$ For the concentration of sulfate anion, $$\frac{0.0166\; mol\:Na_{2}SO_{4}}{1\:L}\;\times\;\frac{1\;mol\;SO_{4}^{2-}}{1\;mol\;Na_{2}SO_{4}}=0.0166\;M\;SO_{4}^{2-}$$ Now that we have calculated the concentrations of each ion in the solution, we can now find the ionic strength of the solution: $$I=\frac{1}{2}\sum_{i=1}^{N}c_iz_i^2=\frac{1}{2}[(0.0332\;M\;Na^{+})(+1)^2+(0.0166\;M\;SO_4^{2-})(-2)^2]=0.0498\;M\approx0.05\;M$$ Since the ionic strength of the solution is approximately 0.05 M, we will refer to that column in the table for activity coefficients. Thus, $$\gamma_{H_3O^+}=0.860$$ $$\gamma_{OH^-}=0.810$$ In order to see how the "spectator" ions of sodium and sulfate affect the pH of the solution, we must consider the ionization (or dissociation) of water and its corresponding ICF chart: $$\begin{tabular}{ l | c | r } 2H2O& H3O+& OH- \\ -&0M&0M \\ -&+x&+x \\ -&xM&xM \\ \end{tabular}$$ Since pH is determined by the concentration of hydronium ions present in the solution, we must find that concentration, and we can do so by using the dissociation constant definition for water: $$K_w=\alpha_{H_3O^+}\cdot \alpha_{OH^-}=\gamma_{H_3O^+}\cdot \gamma_{OH^-}[H_3O^+][OH^-]=(0.860)x(0.810)x=0.6966x^2=1.0\times 10^{14}$$ $$x=[H_3O+]=1.2\times10^{-7}\;M$$ Now that we have the concentration for water, we must find its activity denoted by $$\alpha_{H_3O^+}$$: $$\alpha=\gamma\cdot[H_3O^+]=(0.860)(1.2\times10^{-7})=1.030\times10^{-7}$$ Now, we can finally calculate the pH of the sodium sulfate solution: $$pH=-log\alpha_{H_3O^+}=-log(1.03\times10^{-7})\approx6.987$$

Basic Chemistry

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Question:

UV-Vis spectroscopy refers to an instrumental method where by light in the ultraviolet and visible range is shined onto various substances in order to excite electrons. Suppose a student is conducting a spectroscopic experiment with the organic compound ethanol, and they choose a light of wavelength 418 nm. What is the frequency of the radiation, and does it fall within the visible or ultraviolet spectrum? If it is visible, what is the corresponding color?

Allyson F.

Answer:

In order to calculate the frequency of the radiation or light, we will use the equation that relates frequency and wavelength: $$c=\lambda\cdot\nu$$, where $$c$$ is the speed of light, $$\lambda$$ is the wavelength, and $$\nu$$ is the frequency. $$2.9979\times10^{8}\;m/s=(418\times10^{-9}\;m)\nu$$ $$\nu=7.172\times10^{14}\;Hz$$ This is in the visible spectrum, and the color is violet.

Calculus

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Question:

A rod of length L meters is aligned with the x-axis. The left end of the rod is located at the point (2,0). The charge density in the rod is given by: $$\lambda_{q}(x)=\frac{ln(x)}{x^{4}}$$ C/m. Determine the total charge in the rod.

Allyson F.

Answer:

Imagine that the rod of length L meters is placed on the x-axis. The left end is placed at (2,0). As the rod is of length L, the right end of the rod will be at (2+L,0). Notice the units of the charge density; since we are asked to find the total charge, which has units of Coulombs (C), we can then simply multiply the charge density by the length of the rod, as the density has units of Coulombs per meter (C/m) and the length has units of meters (m). Multiplying these two items together will cause the meters to cancel, just leaving the units of Coulombs. However, we cannot simply multiply the total length, which we will further refer to in terms of the variable "x", by the charge density. This is because the charge density is not uniform, but rather, it is a function with respect to "x", or the length of the rod. This means that the charge density is different depending on where you are on the rod. Thus, in order to figure out the total charge in the rod, we will divide the rod up into tiny pieces, and we will look at the charge in each tiny piece and add them together to find the total charge. The rod will be divided up into an arbitrary number of pieces that we will call N. Imagine the rod being cut up into pieces. Choose a random piece on the rod, and call the left end up the piece $$x_{i}$$. Note that we can label either the right-hand side or the left-hand side as $$x_{i}$$, but for this example, we will choose the left-hand side. The length of this very small piece is $$\Delta x$$. Then, the charge in this piece is: $$\Delta q=\lambda_{q}(x_{i})\Delta x$$. Suppose we do this process for all of the N pieces that we cut the rod into, and then we add them all together to approximate the total charge in the rod. We can represent this as a summation: $$\sum_{i=0}^{N-1}\lambda_{q}(x_{i})\Delta x=\sum_{i=0}^{N-1}\frac{ln(x_{i})}{x_{i}^{4}}\Delta x$$. Notice that the upper limit of the summation is N-1; this is because we chose the left-hand side of the small piece of the rod. If we would have chosen the right-hand side, the lower bound would be "i=1" and the upper bound would be N. To get the best approximation, we will assume that the number of pieces is infinite. This is a classic example of a Riemann sum which is turned into a definite integral. Thus, the total charge is equal to: $$q\approx\lim_{N\rightarrow \infty}\sum_{i=0}^{N-1}\lambda_{q}(x_{i})\Delta x=\lim_{N\rightarrow \infty}\sum_{i=0}^{N-1}\frac{ln(x_{i})}{x_{i}^{4}}\Delta x=\int_{2}^{2+L}\frac{ln(x)}{x^{4}}dx$$ Notice the bounds of the integral; they come from the length of the rod aligned on the x-axis. Now, we must simply solve the definite integral to find the total charge in the rod. In order to solve this integral, we must employ the method of integration by parts. When checking what method to use when integrating, first we try u-substitution, but we cannot use that in this situation. A general rule of thumb when we see ln(x) in the integrand multiplied by something else is to try and use integration by parts. $$q=\int_{2}^{2+L}\frac{ln(x)}{x^{4}}dx$$ $$u=ln(x), u'=\frac{1}{x}$$ $$v'=x^{-4}, v=\frac{-1}{3}x^{-3}$$ $$q=\int_{2}^{2+L}\frac{ln(x)}{x^{4}}dx=\frac{-ln(x)}{3x^{3}}\Big|_2^{2+L}-\int_{2}^{2+L}\frac{-1}{3}x^{3}\cdot x^{-1}dx=\frac{-ln(x)}{3x^{3}}\Big|_2^{2+L}+\frac{1}{3}\int_{2}^{2+L}x^{-4}dx=\frac{-ln(x)}{3x^{3}}\Big|_2^{2+L}+\frac{1}{3}(\frac{-1}{3}x^{-3})\Big|_2^{2+L}=\frac{-ln(2+L)}{3(2+L)^{3}}+\frac{ln2}{3(2)^{3}}-\frac{1}{9(2+L)^{3}}+\frac{1}{9(2)^3}=\frac{ln2}{24}-\frac{ln(2+L)}{3(2+L)^{3}}-\frac{1}{9(2+L)^{3}}+\frac{1}{72}=\frac{3ln2+1}{72}-\frac{3ln(2+L)-1}{9(2+L)^{3}}\; C$$

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