Calculate the pH of water at room temperature? Assumption: the activity coefficients of both OH- and H3O+ are ~1 (given their low concentrations). pH = -log10[aH3O+] . a: activity.
Added by Kurt C.
Step 1
Let's think step by step. Show more…
Show all steps
Your feedback will help us improve your experience
Adi S and 88 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find [H+][OH-] in a solution with a pH of 3.0 at 25°C. You do not need your calculator for this, which means that you could see this sort of thing in the multiple choice section of the AP Exam. What is the hydroxide ion concentration in a solution with a pH of 9.5 at 25°C? The pH of distilled water at 25°C is 7.0. When the temperature increases to 37°C, the pH drops to 6.8. At both temperatures, the water is considered to be neutral as [H+]=[OH-]. Explain why the pH drops when the temperature increases.
Adi S.
In pure water at 25 °C, the concentration of water is 55.5 M—grams of H2O in 1 L divided by its gram molecular weight: (1,000 g/L)/(18.015 g/mol)—and is essentially constant in relation to the very low concentrations of H+ and OH−, namely 1 × 10−7 M. Accordingly, we can substitute 55.5 M in the equilibrium constant expression (Eqn 2–3) to yield Keq = [H+][OH−] ⁄ [55.5 M] On rearranging, this becomes (55.5 M)(Keq) = [H+][OH−] = Kw where Kw designates the product (55.5 M)(Keq), the ion product of water at 25 °C. The value for Keq, determined by electrical-conductivity measurements of pure water, is 1.8 × 10−16 M at 25 °C. Substituting this value for Keq in Equation 2–4 gives the value of the ion product of water: Kw = [H+][OH−] = (55.5 M)(1.8 × 10−16 M) = 1.0 × 10−14 M2 Thus the product [H+][OH−] in aqueous solutions at 25 °C always equals 1 × 10−14 M2. When there are exactly equal concentrations of H+ and OH−, as in pure water, the solution is said to be at neutral pH. At this pH, the concentration of H+ and OH− can be calculated from the ion product of water as follows: Kw = [H+][OH−] = [H+]2 = [OH−]2 Solving for [H+] gives [H+] = −−−√ Kw = −−−−−−−−−√ 1 × 10−14 M2 [H+] = [OH−] = 10−7 M As the ion product of water is constant, whenever [H+] is greater than 1 × 10−7 M, [OH−] must be less than 1 × 10−7 M, and vice versa. When [H+] is very high, as in a solution of hydrochloric acid, [OH−] must be very low. From the ion product of water we can calculate [H+] if we know [OH−], and vice versa.
Shaiju T.
Sri K.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD