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Answer and Explanation: The answer is A. 50.0 mL.30 mL is your answer.What volume of 0.10M H2SO4 must be added to 50ml of a 0.10M NaOH to make a solution in which the molarity of H2SO4 is 0.050M? Answer is 100ml.
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How many mL of a 0.10 M NaOH solution are needed to neutralize 15ml of a 0.20 M h3po4?
30 mL is your answer.
What volume of 0.10 mh so must be added to 50 mL of a 0.10 M NaOH solution to make a solution in which the molarity of the H so is 0.050 m?
What volume of 0.10M H2SO4 must be added to 50ml of a 0.10M NaOH to make a solution in which the molarity of H2SO4 is 0.050M? Answer is 100ml.
What is the volume of 0.1 NaOH?
The volume of 0.1 M NaOH required is 0.4 L or 400 mL.
How will you prepare 0.1 m NaOH solution in 250 mL?
To make 250 ml of 0.1 M NaOH, you dissolve 1 gram NaOH in enough water to make a final volume of 250 mls.
How do you make 0.1 m from 1M?
Take 1 part of your stock solution and add 9 parts of solvent (usually water but sometimes alcohol or other organic solvent). In all cases you are diluting by the same factor. The concentration of the resulting solution is 1M /10 = 0.1M where 10 is the dilution factor.
What volume of a 0.100 M hc1 solution is needed to neutralize 25.0 mL of 0.350 M NaOH?
Calculate the volume of 0.100 M HCI solution needed to neutralize 25.0 mL of 0.350 M NaOH solution. (Answer: 87.5 mL) Calculate the volume of 0.100 M HzSO4 solution needed to neutralize 124 mL of 0.250 M NaOH solution.
How many milliliters of 0.1 M H2SO4 must be added to 50 ml of 0.1 M NaOH to give a solution that has a concentration of 0.05 m in H2SO4?
so, millilitres of H2SO4 added is 100 ml.
What volume of 0.10 m H2SO4 will be required to produce 17g of h2s?
Henc, 25 L of 0.1 M H2SO4 is required.
How many milliliters of 0.1 N H2SO4 will be will be complete reaction with a solution containing 0.125 g of pure na2co3?
So answer is 23.58 mil.
How will you prepare 0.1 M sodium hydroxide solution?
- Take about 100ml of distilled water in a cleaned and dried 1000 ml volumetric flask.
- Add about 4.2 gm of Sodium hydroxide with continues stirring.
- Add more about 700ml of distilled water, mix and allow to cool to room temperature.
How do you find the volume of NaOH?
- 25.00 cm 3 of 0.300 mol/dm 3 sodium hydroxide solution is exactly neutralised by 0.100 mol/dm 3 sulfuric acid. …
- Volume of sodium hydroxide solution = 25.0 ÷ 1000 = 0.0250 dm 3
- Amount of sodium hydroxide = concentration × volume.
- Amount of sodium hydroxide = 0.300 mol/dm 3 × 0.0250 dm 3
- = 0.00750 mol.
How do you find the normality of 0.1 N NaOH?
Normality Calculation of NaOH
To make a 1N solution of NaOH, 40 grams of NaOH are dissolved in 1 L. Likewise, for a 0.1 N solution of NaOH, divide by a factor of 10 and 4 grams of NaOH per liter is needed.
How would you prepare 100 ml of a 0.1 M solution?
To make 100 ml of 0.1M NaCl, one would pipette 10 ml of stock 1M NaCl into a 100 ml volumetric flask, and bring the total volume to 100 ml with water.
How do you make a 0.1 M solution?
To make a 0.1M NaCl solution, you could weigh 5.844g of NaCl and dissolve it in 1 litre of water; OR 0.5844g of NaCl in 100mL of water (see animation below); OR make a 1:10 dilution of a 1M sample.
What volume of 0.40 m’ba oh 2 must be added to 50.0 mL of 0.30 M Naoh to get a solution in which the molarity of the OH ions is 0.50 m?
Approximately 33 mL of the Ba(OH)2 solution needs to be added.
What volume of water is added to a mixture of 250 mL of 0.6 M HCL and 750 mL of 0.2 M HCL to obtain 0.25 M solution?
V = 1.2L. Was this answer helpful?
What volume in mL of 5M H2SO4 should be added to 150 mL of 1m H2SO4?
Henceforth, the volume of a 5M H2 SO4 solution should be added to a 150 mL of 1. M H2SO4 solution to obtain a solution of sulphuric acid of molarity 2.5 is 90 ml.
What volume of 0.1 and hno3 solution can be prepared from 6.3 g of hno3?
Hence, volume required is 1L .
how many ml of a 0.10 m naoh
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How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4? | Wyzant Ask An Expert
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- Summary of article content: Articles about How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4? | Wyzant Ask An Expert Since 1 ml of this base solution contains 0.1 mEq, 9/0.1 =90 ml are needed. … (*)The gram equivalent(E) of a molecule can be calculated as the … …
- Most searched keywords: Whether you are looking for How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4? | Wyzant Ask An Expert Since 1 ml of this base solution contains 0.1 mEq, 9/0.1 =90 ml are needed. … (*)The gram equivalent(E) of a molecule can be calculated as the … How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4?
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How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4 solution? | Socratic
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- Most searched keywords: Whether you are looking for How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4 solution? | Socratic Updating 30 mL Use the formula M_AV_A = M_BV_B. M_A = Molarity of acid, V_A =Volume of acid M_B = Molarity of base, V_B =Volume of base 1. Plug known values into the formula. (.20)(15) = .10(x) 2. Solve for x. 3 = .10x x = 30 30 mL is your answer.
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Calculate the volume of 0.1 M NaOH solution required to neutralise the products formed by
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Mass of NaOH needed to make 250mL of 0.1M NaOH solution? | Wyzant Ask An Expert
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How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4 solution? | Socratic
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- Summary of article content: Articles about How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4 solution? | Socratic 30 mL. Explanation: Use the formula MAVA=MBVB . MA= Molarity of ac, VA= Volume of ac. MB= Molarity of base, VB= Volume of base. …
- Most searched keywords: Whether you are looking for How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H_3PO_4 solution? | Socratic 30 mL. Explanation: Use the formula MAVA=MBVB . MA= Molarity of ac, VA= Volume of ac. MB= Molarity of base, VB= Volume of base. 30 mL Use the formula M_AV_A = M_BV_B. M_A = Molarity of acid, V_A =Volume of acid M_B = Molarity of base, V_B =Volume of base 1. Plug known values into the formula. (.20)(15) = .10(x) 2. Solve for x. 3 = .10x x = 30 30 mL is your answer.
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What volume of 0.10 M H(2)SO(4) must be added to 50 mL of a 0.10 NaOH solution to make a solution in which molarity of the H(2)SO(4) is 0.050M?
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How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4?
Although there are several valid approaches to answering this question ( see other answers), practicing chemists have been using conveniently the following steps:
1. Conversion of all concentrations to normal concentrations ( N , number of gram equivalents/L ) (*)
2. Using the following formulae : Vx Nx = V N (1)
From (1)
Vx = V N/Nx (1a)
and Nx = N V/Vx (1b)
where V ,Vx and N,Nx are the volumes and the normal concentrations of a the known and unknown solutions respectively.
1b has general application in volumetric analysis (e.g. titrations) and 1a is useful in our case and in all preparative calculations for all chemical reactions.
step1. Conversion to normalities: 0.1 M NaOH = 0.1 N NaOH (*)
0.2 M H3PO4 = 0.6 N H3PO4 (*)
Step2. Substituting in 1a above: Vx =0.6/0.1 X 15 =90 ml. Of course if remembering or using formulae is not your way to work, a free thinking alternative is available. In this case also, conversion to normal concentration is indispensable for direct comparisons.
Here we have 15 ml of a solution that is 0.6N this means we have 15X0.6 =9 milliequivalents (mEq) of acid . Hence we will need 9 mEq of a base for a complete neutralization ,and we must get them from a 0.1N NAOH solution. Since 1 ml of this base solution contains 0.1 mEq, 9/0.1 =90 ml are needed.
———————————————————
(*)The gram equivalent(E) of a molecule can be calculated as the ratio of the gram mole (M) and the valence (V) of the compound. E= M/V (2). The valence of a reactant in a neutralization reaction can be calculated as the product of the valence of the group involved in the reaction and the number of groups present in the molecule.E.g. for the acid H2SO4, the group involved in acid-base reactions is the proton H+ that is monovalent. Since there are 2 of them, the valence of this acid is 1X2=2. Sulfuric acid is a divalent acid. Hence E=M/2. And for H3PO4 , E=M/3. Phosphoric acid is a trivalent acid. Of course if the neutralization is only partial the valence is equally reduced.in the case of H3PO4 if only one H is neutralized as in the reaction H3PO4 + NaOH= NaH2PO4 +H2O the valence of the acid is 1 because the number of groups (H+) participating is only one.
Mass of NaOH needed to make 250mL of 0.1M NaOH solution?
0.1 M NaOH means to dissolve 0.1 moles NaOH in 1 liter of solution. Since you don’t want to make 1 liter of solution, but want only 250 ml (0.25 liters), then you also don’t need 0.1 moles NaOH. You need only 1/4 of that. So done by dimensional analysis, it looks like this…
0.1 moles/liter x 0.25 liters = 0.025 moles NaOH needed
The molar mass of NaOH is 23 + 16 + 1 = 40 g/mole
Thus, 0.025 moles NaOH x 40 g/mole = 1 gram NaOH
To make 250 ml of 0.1 M NaOH, you dissolve 1 gram NaOH in enough water to make a final volume of 250 mls.
[Solved] Please help How many mL of a 010 M NaOH solution are
The neutralization reaction between NaOH and H₃PO₄ is as follows:
3NaOH(aq) + H₃PO₄(aq) → Na₃PO₄(aq) + 3H₂O(l)
The molarity of NaOH is given as 0.10 M and the molarity of H₃PO₄ is given as 0.20 M. The volume of H₃PO₄ is given as 15 mL. The molarity (M) of a solution can be defined as the ratio of the total number of moles of the given solute to the volume of that solution in liters. Normality (N) is determined as the ratio of the number of gram equivalents of the corresponding solute to the volume of that solution in liters. The relation between molarity and normality is expressed as follows:
Normality = Molarity × n factor ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 1 ) \text{Normality}=\text{Molarity}\times n\text{ factor }\cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right) Normality = Molarity × n factor ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 1 )
The number of replaceable H⁺ and OH⁻ ions of one mole of acid and base respectively is termed its n-factor. The n-factor of H₃PO₄ and NaOH is 3 and 1 respectively. The volume of NaOH can be calculated using the following equation:
N 1 V 1 = N 2 V 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 2 ) {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right) N 1 V 1 = N 2 V 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 2 )
Where _N_₁ and _V_₁ indicate the normality and volume of H₃PO₄, respectively, and _N_₂ and _V_₂ indicate the normality and volume of NaOH, respectively.
Using equation (1), equation (2) can be written in terms of n-factor as follows:
n 1 M 1 V 1 = n 2 M 2 V 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 3 ) {{n}_{1}}{{M}_{1}}{{V}_{1}}={{n}_{2}}{{M}_{2}}{{V}_{2}}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right) n 1 M 1 V 1 = n 2 M 2 V 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( 3 )
Where _n_₁ and _M_₁ indicate the n-factor and molarity of H₃PO₄, respectively, and _n_₂ and _M_₂ indicate the n-factor and molarity of NaOH, respectively. Substitute the respective values in equation (3) to calculate the volume of NaOH as follows:
3 × 0.20 M × 15.0 mL = 1 × 0.10 M × V 2 \begin{aligned} 3\times 0.20\text{ M}\times \text{15}\text{.0 mL}&=1\times 0.10\text{ M}\times {{V}_{2}} \end{aligned} 3 × 0.20 M × 15 .0 mL = 1 × 0.10 M × V 2
V 2 = 90 mL {{V}_{2}}=90\text{ mL} V 2 = 90 mL
Hence, the volume of NaOH required is 90 mL.
Your question is too long, so unfortunately we cannot answer it all with one single answer. So, here’s just part 1, please submit the next parts as separate questions so I can better explain each of them.
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