If 50 ml of 0.5 M H2SO4 and 150 ml of 0.25 M H2SO4 is added and the mixture is made to 250 ml, what is the final concentration of the solution?
final concentration=M1v1+M2V2/total volumeM1=0.5MV1=50ml/1000 = 0.05LM2=0.25MV2=150ml/1000 = 0.15Lfinal volume = 250ml(given in the question (made to 250ml)) = 250ml/1000 = 0.25by applying the formulafinal concentration = 0.05*0.5+0.15*0.25/0.25so the answer is 0.25Mfinal concentration will be 0.25M
If a 1 ml 1M HCl solution is added to 99ml of H2O at 25°C, then what is the pH change of water equal to?
your are diluting the original 1M Hcl to 100 times (1 to 100) hence concentration decreases by 100 times or 1 M becomes 0.01MApplying -log(H+) we getpH = 2
How many grams of NaOH are needed to make 100 mL of a 0.5 M solution of NaOH?
Talking about Concept involvedAs we know Molarity = no of moles of solute/Volume of solution in litresi.e M = m/V(L)M= Molarity of substancem = number of moles of substance (i.e solute)V(L) = volume of solution in LitresSo talking about our question molarity should be 0.5 M,volume should be 0.1 L , and we should know that Molar Mass of NaOH is 40 because Atomic mass of Sodium i.e (Na known as Natrium) is 23,Atomic mass of Oxygen (O) is 16 and Atomic mass of Hydrogen is 1.Adding them all up gives Molecular mass of NaOH as 40So if we solve above equation we get no of moles of NaOH present in 0.1 litres of solutionLet’s do that mathRearranging equation we get m=M×V(L) i.e m= 0.5×0.1 =0.05. So 0.05 moles of NaOH are present in 100 ml of 0.5 Molar solutionNow we also know that to find no of moles of a substance we have a formula and it is [math]m=W/M_w[/math] where[math]m[/math] = no of moles of a substance[math]W[/math] = Mass of substance[math]M_w[/math] = Molecular weight of the substanceSo substituting all values and rearranging equation we get mass of NaOH as[math]W=m×M_w[/math] = 0.05×40=2 gramsHence we found the required mass of NaOH as 2 grams.Now talking about shortcutwe know as explained above [math]M=m/V(L)[/math] and [math]m= W/M[/math]Substituting value of m(no of moles) in [math]M= m/V(L)[/math] we get [math]M= \frac{W}{M_w × V(L)}[/math]Try substituting the values yourself and get the answerNote :: The value of Volume should always be substituted in terms of Litre only. That is the reason i used the letter (L) at all the places with VSo thats all get the answer and comment any other problems belowHope you got your problem solvedThanks!!!
How much HCL is required to neutralize 0.1n NaOH?
It depends on the volume of 0.1N NaOH to be neutralized & the concentration of HCl used. You can find by using formula N1V1=N2V2. That's equating the milliequivalents of the two substances reacting together.
What is the normality of sulphuric acid?
Calculation of Sulphuric Acid Normality:Sp. Gravity of H2SO4 = 1.84 ; Equivalent weight = 49.04 ; Purity of H2SO4 = 98%Normality of H2SO4 = Specific Gravity of H2SO4 x Purity of H2SO4 x 1000/Equivalent weight of H2SO4 x 100Normality of H2SO4 = 1.84 x 98 x 1000/ 49.04 x 100Normality of H2SO4 = 36.8 N