na+oh +25=53
ph+poh=2.3
You must know the method and not just the answer. Think about this problem: What does 5% mean? It tells you that for every 100g (or ml) water, you must have 5g of NaOH. You have 235 ml water. Therefore you will have to use more NaOH. You can calculate this as follows: (235/100)*5= 11.75g NaOH
On the pH paper Merck no. 9565 blue is acidic (pH=5).
- log(1 X 10^-5 M) = 5 14 - 5 = 9 pH ----------
ph 5 is higher than ph 1
Shampoo typically has a pH range of 4.5 to 6.5, making it slightly acidic to neutral. This pH range is formulated to be close to the natural pH of the scalp, aiding in maintaining healthy hair and scalp conditions.
The pH of a 2.34x10^-5 NaOH solution is 12.33 (calculated as -log[OH^-]). The pOH of the same solution is 1.67 (calculated as -log[NaOH]).
To calculate the pH at the 10 mL point in the titration of 50 mL of 1 M acetic acid with 1 M NaOH (Ka = 1.75 x 10^-5), we use the Henderson-Hasselbalch equation. At this point, the acetic acid has been partially neutralized by the NaOH. The moles of acetic acid remaining can be calculated using the initial volume and concentration of acetic acid and the volume of NaOH added. Then, use these values in the Henderson-Hasselbalch equation to determine the pH.
Strongest Acid is HCl. Then in order of acidic strength H2CO3 , NH3(solution) and finally NaOH. In words, Hydrochloric Acid ; pH = 1 Carbomic Acid ; pH ~ 5 Ammonia solutioon ; pH ~ 8 or 9 Sodium Hydroxide ; pH ~ 12
You must know the method and not just the answer. Think about this problem: What does 5% mean? It tells you that for every 100g (or ml) water, you must have 5g of NaOH. You have 235 ml water. Therefore you will have to use more NaOH. You can calculate this as follows: (235/100)*5= 11.75g NaOH
pH depends on concentration. Strong bases are for example: NaOH KOH Ca(OH)2 http://en.wikipedia.org/wiki/Base_(chemistry)
5 M NaOH = 5 mol/L NaOH. 1 mol of NaOH = 40 g NaOH. You can get the number of grams by multiplying 5 mol/L and 40g/mol. This will give you 200g/L. Just multiply by the volume, which is 1 Liter, and you will get 200 g of NaOH.
pH=-log[H+] pH=-log[1.00x10^-5M] pH=5 When concentration of H+ is 1M, the pH is zero. Every time the concentration is decreased by a magnitude of 10 (i.e. 10^1 ---> 10^-2) the pH goes up 1 value. This is true for the pH of bases as well, but in that case pH=14+log[OH-], which is derived from pOH=-log[OH] and pH+pOH=14. This is true for pure water at 25 degrees Celsius.
Well, isn't that a happy little question! To prepare 0.25 L of 0.5 M NaOH solution, you'll need to dissolve the correct amount of NaOH in water. You can use the formula C1V1 = C2V2 to calculate the amount of NaOH needed. Just take your time, follow the steps gently, and you'll have a perfect solution in no time.
No is the answer. Sodium hydroxide is is such a strong alkali that it will 'burn' the cells just like acid does. So it is not found in body. Instead you have sodium bi carbonate. But you have very strong acid in stomach. Hydrochloric acid. Imagine acidity go on increasing ten times as you go from 6 to 5 and 5 to 4. And pH in stomach is 1 to 2. (Why does not stomach is 'not' charred at this pH ?) Pepsin works in high acidic pH only and almost stop working at 5.5 pH. So no question of digestion in sodium hydroxide, which is strong alkali with very high pH. If you lower the pH pepsin starts working again.
1. Weigh 20 g NaOH. 2. Put this NaOH in a 1 L volumetric flask. 3. Add slowly 200 mL distilled water and stir. 4. Put the flask in a thermostat at 20 0C and maintain for 1 hour. 5. Add distilled water up to the mark. Stir vigorously. 6. Standardize the solution by titration with oxalic acid, potassium hydrogen phtalate, etc. 7. Transfer the solution in a bottle and apply a label (date, name of the operator, name of the solution, normality).
On the pH paper Merck no. 9565 blue is acidic (pH=5).
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution