Question

1. calculate the ph of each solution after the addition of 0.010 mol hcl to each of the following solutions. (assume no change in volume)

a. 250.0 ml pure water
b. 250.0 ml 0.10 m naoh
c. 250.0 ml of a buffer that contains 0.255 m nh₃ and 0.245 m nh₄cl

1. the kₐ of a certain indicator is 2.0 x 10⁻⁶. the color of hin is green and that of in⁻ is red. a few drops of the indicator are added to a hcl solution which is then titrated against a naoh solution. at what ph will the indicator change color?
2. methyl red has a pkₐₙ of 4.95. it changes from red to yellow over the ph range from 4.4 to 6.2. indicate the color of each of the following solutions upon the addition of methyl red.

a. 0.100 m hcl b. 0.100 m naoh c. 0.100 m nacl
d. 0.100 m nh₄cl e. 0.100 m nh₃ f. 0.100 m ch₃cooh

Explanation:

a. 250.0 mL pure water

Step1: Calculate the molarity of HCl

The moles of HCl is 0.010 mol and the volume is 0.250 L. The molarity $M=\frac{n}{V}$, so $M_{HCl}=\frac{0.010\ mol}{0.250\ L}=0.04\ M$. Since HCl is a strong - acid and completely dissociates, $[H^+]=0.04\ M$.

Step2: Calculate the pH

The formula for pH is $pH = -\log[H^+]$. So, $pH=-\log(0.04)= 1.40$.

Step1: Calculate the moles of NaOH

$n_{NaOH}=M\times V = 0.10\ M\times0.250\ L = 0.025\ mol$.

Step2: Determine the moles of HCl and NaOH after reaction

HCl reacts with NaOH in a 1:1 ratio. Moles of HCl = 0.010 mol. Moles of NaOH remaining after reaction $n = 0.025\ mol-0.010\ mol = 0.015\ mol$.

Step3: Calculate the molarity of $OH^-$

$[OH^-]=\frac{0.015\ mol}{0.250\ L}=0.06\ M$.

Step4: Calculate the pOH and then pH

$pOH = -\log[OH^-]=-\log(0.06)=1.22$. Since $pH + pOH=14$, then $pH = 14 - 1.22=12.78$.

Step1: Calculate the moles of $NH_3$ and $NH_4^+$ initially

$n_{NH_3}=M\times V=0.255\ M\times0.250\ L = 0.06375\ mol$, $n_{NH_4^+}=0.245\ M\times0.250\ L = 0.06125\ mol$.

Step2: Determine the moles of $NH_3$ and $NH_4^+$ after reaction with HCl

HCl reacts with $NH_3$ to form $NH_4^+$. Moles of $NH_3$ after reaction $n_{NH_3}=0.06375\ mol - 0.010\ mol=0.05375\ mol$. Moles of $NH_4^+$ after reaction $n_{NH_4^+}=0.06125\ mol+0.010\ mol = 0.07125\ mol$.

Step3: Use the Henderson - Hasselbalch equation

The $pK_b$ of $NH_3$ is 4.75, so $pK_a = 14 - 4.75 = 9.25$. The Henderson - Hasselbalch equation is $pH=pK_a+\log\frac{[A^-]}{[HA]}$, where $A^-$ is $NH_3$ and $HA$ is $NH_4^+$. $pH = 9.25+\log\frac{0.05375\ mol/0.250\ L}{0.07125\ mol/0.250\ L}=9.25+\log\frac{0.05375}{0.07125}=9.13$.

Answer:

1.40

b. 250.0 mL 0.10 M NaOH