Adding 100. ml of 0.100 M NaOH to 100. ml of 0.175 M NH4Cl, calculate the pH for the solution. (Ka for NH4+ is 5.70x10-10) (Use ICE table)

Q: 1- Calculate the pH of 2 solution prepared by adding 25 ml of 0.IM of NAOH solu. to 30 ml of 0.2M of…

A: "Since you have posted multiple questions, we will solve only the first question for you according…

Q: A chemist titrates 50.0 mL of a 0.5648M potassium hydroxide (KOH) solution with 0.2574M HCl solution…

A: we have to calculate pH at equivalence point foe the titration

Q: he pH of a solution formed by mixing 20.0 mL of 0.250 M formic acid, HCOOH, and 8.0 mL of 0.400 M…

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Q: A solution is prepared that is initially 0.39M in hypochlorous acid (HC10) and 0.28M in sodium…

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Q: Calculate pH for a solution that contains proton [H+] of 0.02 mM (milimole/L).

A: Given: The concentration of proton [H+] is 0.02 mM

Q: Calculate the pH when 15.52 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

A: The neutralization reaction is given below.

Q: For 280.0 mL of pure water, calculate the initial pH and the final pH after adding 0.024 mol of NaOH

A: pure water : pH = 7.00 moles NaOH = 0.024 [OH-]= 0.024 mol/ 0.280 L=0.0857 M pOH = - log 0.0857=…

Q: A solution is prepared that is initially 0.39M in pyridine (C,H,N), a weak base, and 0.058M in…

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Q: Find the pH when 20.00 mL of 0.087 42 M NaOH have been added to 25.00 mL of 0.066 66 M HClO4.

A: We know that, pH and pOH can be expressed as the negative logarithm of H+ and OH- respectively ,…

Q: What is the pH when 29.3 mL of 0.10 M NaOH is added to 25.0 mL of 0.35 M HF?

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Q: A chemist makes a new solution: 1.0 L of a solution that is both that is both 0.6 M NH3 and 0.8 M…

A: Henderson Hasselbach equation is used to calculate the pH of a buffer system .  pH=pKa +…

Q: Kb for 0.100 M phenol, whose pH is 8.82 is

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Q: Calculate the pH when 19.15 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

A: The neutralization reaction is given below.

Q: Calculate the pH of a solution obtained by mixing 25.00 mL of 0.20 M NH3 with 25.00 mL of 5.0 x 10-2…

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Q: You have 500mL of a solution 0.3 M acetic acid (Ka= 1.74x10^-5 M). The pH of the solution before and…

A: We have to calculate the pH after addition of base.

Q: Calculate the pH of a solution prepared by mixing 10.0 mL of 0.3 M HCl with 15.0 mL of 0.5 M NaOH.

A: We have given that Concentration of HCl = M1 = 0.3 M Volume of HCl = V1 = 10 mL Concentration of…

Q: Determine the pH if a solution that consists of 175.0mL of 0.085M Benzoic acid and 100.0mL of 0.045M…

A: Given information: Ka of benzoic acid = 6.3 × 10-5. Benzoic acid = 175 mL of 0.085 M Sodium benzoate…

Q: Calculate the pH when 9.06 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

A: From the molar equation-   M1: molarity of NaOH solution. V1: volume of NaOH solution. M2: molarity…

Q: Determine the pH of the resulting solution when the following two solutions are mixed: 40.0 mL of…

A: In this question, we have to find out the correct answer of given problem by the help of the…

Q: Pls help ASAP. Pls show all work and calculations.

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Q: A 1000.0 mL solution contains 0.150 M NaCIO and 0.200 M HCIO. What is the pH of this solution before…

A: pH = pKa + log[salt]/[acid] pKa = -logKa

Q: 16. If 625 mL of 2.00 M sodium hydroxide is mixed with 50.0 mL of 10.0 M sulfuric acid, will the pH…

A: The pH of any compound can be calculated on the basis of the concentration of hydronium ions present…

Q: Calculate the pH when 4.91 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

A: The volume of NaOH solution is 4.91 mL which is equal to 0.00491 L and the volume of acetic acid is…

Q: Show the pH calculation process when 24.00, 25.0, 30.00 mL of 0.100 M NaOH are added to 50.00-mL…

A: HCl and NaOH react with each other to form water and NaCl salt.  In any aqueous solution the…

Q: Calculate the pH after 75.0 mL of 0.100 mol L- HCl solution has been added to 100.0 mL of 0.100 mol…

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Q: Calculate the pH when 5.48 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

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Q: Calculate the pH of a 0.50 M HCOOH/ 0.60 M HCOOK buffer solution, and calculate the pH when 5.00 mL…

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Q: What is the pH of the solution after 50.00 mL of 0.100 M  NaOH  is added to 50.00 mL of 0.100 M…

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Q: A solution is prepared that is initially 0.095M in methylamine (CH3 NH,), a weak base, and 0.38M in…

A: we need to fill the given table and calculate pH of the solution

Q: 5. A 1.50M acetic acid solution reacts with 20.0 mL 0.200M NaOH. If 20.0 mL of acid is used,…

A:

Q: Pls help ASAP. Pls show all work and calculations.

A: Balanced Chmical equation for the reaction is as follows: HCl + NaOH ------> NaCl + H2O Both HCl…

Q: A solution is prepared that is initially 0.24M in ammonia (NH,), a weak base, and 0.28M in ammonium…

A: In a chemical reaction, the state in which both reactants and products are present at…

Q: A general chemistry student mixes together a solution of HF with a solution of NaF. If the resulting…

A: Acidic buffer solution :- A mixture of weak acid and its salt with strong base is called acidic…

Q: • Calculate the pH after the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCI.

A: Given :  Concentration of HCl= 0.100 M  Volume of HCl = 25.0 ml  Concentration of NaOH= 0.100 M…

Q: a) Consider CH3COONH4: What is the solution pH if this compound is dissolved in it? Given the Ka of…

A: Given,  Acid dissociation constant of CH3COOH = Ka = 1.8 x 10-5  Acid dissociation constant of NH4+…

Q: A chemist titrates 250.0 mL of a 0.0775 M potassium hydroxide (KOH) solution with 0.0631 M HCl…

A: Given is an acid- base titration, where base is KOH and acid is HCl. Here we have to find the ph at…

Q: Calculate the pH of a solution obtained by mixing 438 mL of 0.14 M  nitric acid with 385 mL of…

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Q: An aqueous solution contains 0.485 M acetic acid. Calculate the pH of the solution after the…

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Q: Calculate the pH of a solution containing 200ml of 0.01M NaOH and 8ml of 0.25M of acetic acid.

A: Answer :-  The pH of a solution containing 200ml of 0.01M NaOH and 8ml of 0.25M of acetic acid =…

Q: Calculate the pH when 9.71 mL of 0.100M NaOH has been added to the original 30.0mL of 0.100M Acetic…

A: Given: Volume of NaOH = 9.71 mL Concentration of NaOH  = 0.100 M Volume of CH3COOH = 30.0 mL…

Q: how much grams of NaNo2 we needed to add into the 100 ml HNO2 solution has a concentration = 0.1…

A: We have to calculate the mass of NaNO2 required.

Q: what is the pH of a solution made by mixing 30 mL of 0.100 M HCl with 40.0 mL of 0.100 M KOH? Assume…

A: Given  Volume of HCl ( V1 )    = 30 mL  Molarity of HCl ( M1 )  = 0.100 M  Volume of KOH ( V2 )   =…

Q: A chemist titrates 250.0 mL of a 0.6994 M hydrobromic acid (HBr) solution with 0.8110 M NaOH…

A: HBr is a strong monoprotic acid. NaOH is also a strong monoacidic base.

Q: Calculate the pH change by taking 40 mL of 0.1M HCl with adding a 0.1M NaOH of 50 mL of the base

A: Given Values: Initial solution= 0.1 M HCl with a volume of 40 ml Now, 50 ml of 0.1 M NaOH is added…

Q: A chemist makes a new solution: 1.0 L of a solution that is both that is both 0.6 M NH, and 0.8 M…

A: NH3 and NH4Cl make a basic buffer solution because NH3 is weak base and NH4Cl is strong acid salt

Q: When 15.00 mL of 0.0500 M HBr is added to 7.50 mL of 0.100 M LIOH, what is the pH of the resulting…

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Q: A chemist titrates 150.0 mL of a 0.8461M hydrobromic acid (HBr) solution with 0.2857M NaOH solution…

A: Given that : The reaction for the process is : HBr(aq)  +  NaOH(aq)  →  NaBr(aq)  +  H2O(l)

Q: Calculate the pH of a 1L aqueous solution that is 0.13 M in NH3 and 0.26 M in NH4NO3 after 0.013 mol…

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Q: An aqueous solution contains 0.368 M hydrocyanic acid. Calculate the pH of the solution after the…

A: When a strong base is added to weak acid a buffer is formed. The pH of the buffer can be calculated…

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.