pH = 14 – pOH You can also calculate pOH in the same manner: pOH = 14 – pH As you can see from these formulas, the higher the concentration of hydrogen ions, the lower the pH will be, and the more acidic the solution will be. On the other hand, the higher the concentration of...
pH and pKa: Both pH and pKa have formulas that can help show how acidic a substance is. pH has its own range, as does pKa. They have a relationship with one another that also can show how acidic something is. Answer and Explanation:1 ...
Use the formulas for pH and pOH to perform the following calculations. 4. 5. What is the pH of 6.6 × 10-4M HCl? 6. What is the pH of 1.5 × 10-3M NaOH? 7. A solution of HNO3 has a pH of 4.5. What is the molarity of HNO3? 8. What is the molarity of KOH in a sol...
\(\begin{array}{l}-log~[OH^-] = pOH\ and -logK_b = pKb\end{array} \) , The above equation can be written as, \(\begin{array}{l}pK_b = pOH ~- ~log\frac{[BH^+]}{[B]}\end{array} \) Rearranging the equation,
There are more formulas and complex mathematical operations of pH calculation in solution of Analytical Chemistry, which are difficult for students to understand and remember. According to the character and individuality of different solutions for pH calculation, the teaching methods of induction, analogy...
All you need to do is solve for x and take the -log of it to find the pH. « Re:pH/Ka «Reply #3 on: First you would need to calculate the pOH, which is the same thing as pH only the concentration of OH-is used in place of the H+concentration. Once you have the pOH,...
pOH=4.77 pH=pKw-pOH=14.00-4.77=9.23 案例3:计算0.20 mol/L邻苯二甲酸溶液的pH。 解析:邻苯二甲酸Ka1=1.12×10-3,Ka2=23.90×10-6。 pH=1.83 (5) 缓冲溶液的缓冲容量与缓冲溶液的总浓度(cHA+cA-)和缓冲比(cA-/cHA)有关。缓冲溶液的总浓度越大,缓冲容量越大;缓冲溶液的缓冲比越接近1,缓冲容量越大...
In this study, changing trends in regional spatial patterns of PHCEs are evaluated by Global Moran's I and Local Indicators of Spatial Association (LISA) by using ArcGIS10.3 and GeoDa software, respectively. The formulas of Global Moran's I are shown below [49]: I = n ∑in=1 ∑nj=i ...
The level of chlorophyll a, chlorophyll b, and carotenoid content of the leaves were calculated using the following formulas [24]: Chlorophyll a = 12.7 × Absorbance at 663 nm − 2.69 × Absorbance at 645 (mM concentration) (1) Chlorophyll b = 22.9 × Absorbance at 645 nm − 4.68 ×...