hi this is a science Chef in this video you'll be learning about the heat of neutralization and how to solve Associated problems on heat of neutralization all right let's start [Music] to calculate the standard height of neutralization of a reaction we apply the formula Delta H theta equals to minus Q Over N where Q is the amount of heat released neutralization reaction between the acid and the base n is the number of moles of either the acid or the base obtained by multiplying their molar concentrations and volumes in dmq of course the negative sign shows its exothermic value as usual to evaluate Q we use Q equals to Ms times CS times delta T where Ms is the mass of the solution mixture obtained by adding the volume of the acid and the base to get the volume of the solution then multiplying by the density of the solution in grams per CM Cube CS is a specific capacity of the solution and delta T is a change in temperature calculated as T2 minus T1 T1 is always the average of the initial temperatures of the acid and the base Solutions while T2 is the final temperature of the mixture let's now see how to apply this formula in solving problems on heat of neutralization example one 75 cm cube of 0.2 molar sodium hydroxide was added to 75 cm cube of 0.2 molar hydrochloric acid in a plastic Beaker of negligible heat capacity the initial temperature of the mixture was 14.7 degrees Celsius this Rose to 16.0 degrees Celsius calculate the standard enthalpy of neutralization given that the specific heat capacity of the solution is 4.2 joules per gram per degree celsius solution one in this question we are given the volumes and concentrations of the acid and the base the initial and final temperatures of the reaction mixture and the specific heat capacity of the solution we also given the density of the solution and we are asked to calculate the enthalpy of neutralization for the information provided we'll calculate the volume of the solution and invariably the mass of the solution using that will determine the value of Q and also calculate the number of moles of the acid or base and thereafter we will use these variables to calculate the standard enthalpy of neutronization of the reaction all right let's go so volume of solution vs equals VA plus VB which is 75 CM Cube plus 75 cmq it gives us 150 CM Cube good but the density of solution is one gram per CM Cube therefore mass of solution equals volume of solution times density of solution which gives us 150 cmq times one gram per CM Cube so that equals 150 grams good so now going to calculate its values into Qs equals Ms times CS times change in Theta and that will be 150 grams times 4.2 joules per grams per resources times change in temperature will be 16 minus 14.7 which is 1.3 degrees Celsius 1.3 degrees Celsius so this and this should go this and this would go so evaluate this we'll get 819 Jews as the heat released by reaction mixture it was used to raise the temperature of the solution then for concentration in most by dmq and volume in dmq we are going to calculate the number of moles of the acid or base as follows number of moles of acid we go to concentration of acid times volume of acid so construction of acid that was given to us 0.2 moles per dmq we are given 0.2 molars molar is the same thing as moles per dmq times 75 cm Cube 75 cm Cube divided by 1000 will give us 75 you got the value in D and Q so evaluate this that will give us 0.015 more if you're now going to divide the hits released by the solution by this number of moles to get the standard entropy of neutralization it should be Delta h n Theta is equal to minus Q solution over n so this would be now be 8 1 9 joules divided by 0.015 [Music] thank you so to evaluate this that will give us remember this will be negative minus 54. 600 joules per month which is approximately minus 54.6 kilojoules per mole so this gives us the value of the enthalpy of neutralization for that particular reaction yes it may not be exactly the same value because minus 7.1 kilojoules per mole is theoretical value bet in terms of the Practical value it may be less it may be what higher there will be a slight worth division all right so let's move on to the next question example two 50 cm cube of 0.33 mole per dmq potassium hydroxide was added to 50 cm cube of 0.33 mole per DM Cube trouser nitrate five acid solution in a caloric matter the initial temperatures of the two solutions were 31 degrees Celsius and 18 degrees Celsius respectively the temperature Rose to 34.1 degrees Celsius given the heat capacity of the calorimeter was 0.488 joules per Kelvin calculate the standard enthalpy of neutralization given that the specific heat capacity of the solution is 4.208 joules per gram per Kelvin solution two here we are given the volumes and concentrations of the acid and the base the specific capacity of the solution the final temperature of the solution mixture and the density of the solution mixture but unlike example one we are given the initial temperatures of the acid and the base and the heat capacity of the calorimeter so to get initial temperature of the mixture we would find the average of the initial temperatures of the acid and the base that is T1 will be equal to TA plus TB all over 2 which is 33 plus 31 divided by 2 and this should give us 32 degrees Celsius note that heat capacity of the calorimeter is a product of its mass and specific heat capacity and it is given in joules per Kelvin this is not a problem at all as well solving problems on heat capacity one Kelvin is always equal to one degree Celsius so there is no need for conversion so let's continue so the volume of solution vs we go to VA plus VB which equals 50 plus 50 that gives us 100 cm Cube but since the density of solution is one gram per CM Cube this implies that the mass of solution will be equal to volume of solution times density of solution which will be 100 cm Cube times 1 gram per cm cubed and that gives us 100 grams so the mass of the solution will be 100 grams so I register to the values of the parameters into the equation QRS equals Ms CS delta T plus MC CC delta T it gives us Ms CS plus MC CC delta T so it's equal to 100 grams times 4.208 joules per gram per degree celsius plus remember the mass of the parameter times the specific capacity of the calorimeter equals to the heat capacity of the calorimeter which is giving us 0.488 joules per degree celsius since one carry the same thing as one degree Celsius okay so this will be in bracket then we multiply everything here by the change in temperature which will be 2.1 degrees Celsius to evaluate that that should give us 884 705 joules as they hit released by reaction mixture which was used to raise the temperature of the solution therefore concentration in most per dm2 times volume in GM Cube we are going to calculate the number of moles of the acid or the base this time let's calculate the number of moles of the base that will be concentration of the base times volume of the base so this gives us 0.33 moles by dmq times 50 over 1000 dmq so this gave us 0.0165 [Music] more please note that it's not every time that your volume of acid and base would be equal not the concentration of acid and base will be equal they can be different but the same principles will follow take note of that therefore standard heat of neutralization will not be equal to minus Q over what n so substitute the values that will be 884 705 joules over 0.0165 more of course negative will obtain 53 000 618 points four eight joules is approximately 53 .618 kilojoules per mole yeah we are leaving answers in three decimal places because the specific capacity of the solution and the heat capacity of the calorimeter that were given to us were given to us in three decimal world places so our answer must also be what three decimal places all right so let's move to the last question example three [Music]