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3 years ago

Solid lead(II) sulfide reacts with aqueous hydrochloric acid to form solid lead(II) chloride and dihydrogen sulfide gas. Express

your answer as a chemical equation.

Chemistry

1 answer:

Vikentia [17]3 years ago

6 0

$PbS(s) + 2HCl(aq) \rightarrow PbCl_2(s) + H_2S(g)$

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Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =

ValentinkaMS [17]

Answer : The concentration of $OH^-$ ion, pH and pOH of solution is, $1.12\times 10^{-13}M$ , 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of $H^+$ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.090)$

$pH=1.05$

The pH of the solution is, 1.05

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95$

The pOH of the solution is, 12.95

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12.95=-\log [OH^-]$

$[OH^-]=1.12\times 10^{-13}M$

The $OH^-$ concentration is, $1.12\times 10^{-13}M$

5 0

2 years ago

Read 2 more answers

What element used to be mined in the upper peninsula of Michigan

Afina-wow [57]

Coal, they use to mine for it alot

4 0

3 years ago

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important?

kati45 [8]

Answer:

Answer: What can experiments in a lab tell us about substances on Titan? Experiments in a lab can tell us that the lake did not evaporate in 2007 because the molecular attraction was a lot stronger, then it got weaker overtime.

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important? Their research differs because they were both talking about different things, Hayes was talking about how many lakes there were, while Malaska's was doing more hands on stuff like experiments. Both are important because we need to learn how the lakes formed, but we also need to do hands on experiments.

Explanation:

6 0

2 years ago

We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of

Arte-miy333 [17]

Answer:

A. pH using molar concentrations = 2.56

B. pH using activities = 2.46

C. pH of mixture = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

HA + H₂O ⇌ A⁻ + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}$

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

$\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}$

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is

$I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041$

(iii) Calculate the activity coefficients

$\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79$

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\$

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

$I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10$