pH Curves for Titrations
I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):

Let's consider the case of  the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:


The pH increases slowly at first because the pH scale is 
logarithmic, which means that a pH of 1 will have 10 times the 
hydronium ion concentration than a pH of 2. Thus, as the 
hydronium ion is initally removed, it takes a lot of base to change 
its concentration by a factor of 10, but as more and more 
hydronium ion is removed, less base is required to change its 
concentration by a factor of 10. Near the equivalence point, a 
change of a factor of 10 occurs very quickly, which is why the 
graph is extremely steep at this point. As the hydronium ion 
concentration becomes very low, it will again take a lot of base to 
increase the hydroxide ion concentration by 10 fold to change the 
pH significantly. 
Let's find the equation of the titration curve in the the region $0&lt;V_b&lt;V_b(eq)$:


The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$

The number of moles of $\ce{H3O+}$ after adding the volume   $V_b$ of the base is $C_aV_a-C_bV_b$

Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
So, the titration curve in the region $0&lt;V_b&lt;V_b(eq)$  $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.