Calculate the ph of 0.01 m naoh? (2023)

What is the pH of 0.01 M of NaOH solution?

If 0.01 mole of NaOH is added to 1 litre solution of a basic buffer having pH=8.7 initially, the pH changes to 9.

(Video) The pH of a `0.001 M` aqueous solution of sodium hydroxide will be

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What is the pH of a .010 M NaOH solution?

10 M NaOH solution corresponds to pH of 15.

(Video) Calculate the pH of a 0.01 M of NaOH solution. | 12 | ALCOHOLS, PHENOLS AND ETHERS | CHEMISTRY ...

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How do you calculate the pH of 0.1 m NaOH?

2. Calculate the pH of a 0.1 M NaOH solution. Alternatively we can calculate pOH = - log [OH^-]= - log [10^-¹] = 1 and remembering that pH + pOH = 14 we get pH = 14 - 1 = 13. 3.

(Video) The pH of 0.01 M NaOH(aq) Solution will be

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What is the pH of 0.01 m?

Hence, the answer is – option (d) – the pH of 0.01 M solution of HCl is 2. =>pOH = 14 – pH. Note: “pH is a scale used to specify the acidity or basicity of an aqueous solution.

(Video) Calculate the pH of 0.2 M NaOH

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What is the pH of 0.01 M NaOH assuming complete ionisation?

∴pH=12.

(Video) Teaching how to Calculate the pH and pOH of 0.01M solution NaOH

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What is the pH of 0.01 m NaCl?

Therefore, NaCl is a neutral salt i.e. there is no acidic or basic character in the salt when dissolved in water. It doesn't change the pH of the solution. Hence it will remains neutral i.e., pH = 7.

(Video) pH, pOH, H3O+, OH-, Kw, Ka, Kb, pKa, and pKb Basic Calculations -Acids and Bases Chemistry Problems

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What is the pH of 0.02 m NaOH?

pOH=−log10(OH)−=−log10(0.02)pOH=1.70pH+pOH=14⇒pH=14−1.70=12.3pH=12.3.

(Video) class 11-Ionic equilibrium.How to find PH of strong acid and base.(NEET/JEE)

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What is the pH of 0.05 m NaOH solution?

1 Answer. The pH is about 12.7.

(Video) Calculate the change in pH of water when `0.01` mole of NaOH are added in 10 litre water.

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What is the pH of 0.2 m NaOH?

(pH=13.30)

(Video) The pH of 0.01 M NaOH(aq) Solution will be

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What is the pH of 0.010 NaOH?

The pH of 0.010 M NaOH is about 12.

(Video) Calculate pH for: (a) `0.001 NaOH`, (b) `0.01N Ca(OH)_(2)`, (c ) `0.01M Ca(OH)_(2)`, (d) `10^(-8

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