Answer:
The compound you will use is the Dibasic phosphate
Explanation:
Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,
In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.
To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.
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The [OH⁻] of the solution is 5.37×10⁵ M.
<h3 /><h3>What is pOH?</h3>
This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.
To calculate the hydroxy ion [OH⁻] concentration we use the formula below.
Note:
- pOH = 14-pH
- pOH = 14-9.77
- pOH = 4.27
Formula:
- [OH⁻] = 1/
................. Equation 1
Given:
Substitute the value into equation 1
- [OH⁻] = 1/
- [OH⁻] = 5.37×10⁵
Hence, The [OH⁻] of the solution is 5.37×10⁵ M.
Learn more about hydroxy ion concentration here: brainly.com/question/17090407
Answer:
D.
Explanation:
-log(1.0x10^-5) = pH
pH + pOH = 14 (rearrange it)
OH- = 10^-pOH = 1.0 x 10^-9
- Hope that helped! Let me know if you need further explantion.
