Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.
Find the value of the fraction.Find the value of the fraction.Find the value of the fraction.Find the value of the fraction.
the value of 6 is 60 000 the value of 7 is 7000 the value of 2 is 200 the value of 4 is 40 the value of 1 is 1
The value is 300.
The answer depends on the value of WHAT! The value of your degree education or the value of your student loan debt!
I will assume you are asking about the pH of pure water if pKw is 14.26. The relationship between pH, pOH, and pKw is as follows: pH + pOH = pKw. If it is pure, neutral water (no acids or bases present), then pH = pOH, so: pH + pOH = 14.26 2(pH) = 14.26 pH = pOH = 7.13
there is not
red car = rotes Auto red car = roter Wagen red car = roter PKW
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Here is the first step to convincing you that the pKa of water should not be 7: water is clearly a weaker acid than hydrogen sulfide, whose pKa is 7.04, or boric acid, 9.14, or hydrogen cyanide, 9.31 The real problem is why the pKa should not be 14.0 -- the same as pKw Here is the reason: Kw is defined as [H+]*[OH-], and this turns out to be 1.E-14. For any acid HA in water, though, Ka is defined as [H+]*[A-]/[HA] So if we think of water as an acid, Ka = [H+] * [OH-] / [H2O] = 1.0E-14 / [H2O] What do we do for [H2O]? Well, 1 g of water occupies 1 mL, near enough, which means that 18 gram of water = 1 mol of water occupies 18 mL. So we can fit 55.6 mol of pure water in 1.00 L. It is therefore fair to say that the "concentration" of water is 55.6 M, and so Ka = 1.8E-16 The log of 1.8E-16 to base 10 is -15.75, which is the pKa value quoted in your question.
In general, at 25°C the sum of pKa and pKb for a conjugate acid-base pair will equal 14 due to the auto-ionization of water. However, at different temperatures, the sum may vary slightly due to changes in the equilibrium constant of water ionization.
To find the pH value using pKa, you first need to identify the acid or base equilibrium involved in the reaction. Then, you can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the conjugate base concentration and [HA] is the acid concentration. By knowing the pKa value and the concentrations of the acid and its conjugate base, you can calculate the pH of the solution.
To find the pH of a solution of NaCN, you first need to calculate the concentration of CN- ions (0.028 M) using the molarity of NaCN. Then, calculate the concentration of H+ ions that CN- will react with. Finally, use the Ka value to find the concentration of H+ ions and convert it to pH using the formula pH = -log[H+]. The resulting pH is approximately 10.88 in this case.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.