At a certain temperature, the ph of a neutral solution is 7.43. what is the value of kw at that temperature? express your answer numerically using two significant figures.

The formula for Kw is:

Kw= [H]*[OH]

where [H] and [OH] are concentrations 

We are given the pH so we can calculate the [H] concentration from the formula:
pH=-log[H] 
7.43 = -log[H]

[H] = 3.715x10^-8 M

In neutral solution, [H] = [OH], therefore: 

[OH]= 3.715x10^-8 M

Calculating Kw:

Kw = (3.715x10^-8)*( 3.715x10^-8)

Kw = 1.38 x 10^-15 

The value of the ionic product of water at neutral pH of 7.43 has been  1.38 .

has the ionic product of water and varies at different temperatures.

The ionic product can be given as:

The hydrogen ion concentration at pH 7.43 will be:

pH = -log [Hydrogen ion concentration]

7.43 =  -log [Hydrogen ion concentration]

Hydrogen ion concentration = 3.715 M

At neutral pH, the hydrogen ion concentration is equal to hydroxide ion concentration.

= 3.715   3.715 M

= 1.38 .

The value of the ionic product of water at neutral pH of 7.43 has been  1.38 .

For more information about ionic product, refer to the link:

link

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"